HDU 1878 欧拉回路
最简单的欧拉回路了，如果结点的出度入度之和不是2的倍数，那么就不是欧拉回路。注意要判断图是否连通。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; int g[MAXN]; int n, m; int f[MAXN]; int h[MAXN]; int r[MAXN]; vector<int> vs[MAXN]; void init() { clr(g); clr(h); rep(i,0,n) f[i] = i, vs[i].clear(); } int find(int x) { return f[x] == x ? x : (f[x] = find(f[x])); } void union_set(int a, int b) { int x = find(a); int y = find(b); f[x] = y; } int main() { int a, b; while(RII(n, m) != EOF) { init(); rep(i,1,m) { RII(a, b); g[a]++; g[b]++; union_set(a, b); } rep(i,1,n) find(i); int k = 0; rep(i,1,n) { if(!h[f[i]]) { h[f[i]] = ++k; r[k] = f[i]; vs[k].PB(i); } else vs[h[f[i]]].PB(i); } int cnt = 0; rep(i,1,k) { int tmp = 0; int num = vs[i].size(); rep(j,0,num-1) tmp += g[vs[i][j]]&1; if(tmp == 0 && num != 1) cnt++; else cnt += tmp / 2; } printf("%d ", cnt); } return 0; }
HDU 3018 Ant Trip
对于本题的图，可能存在多个连通分量，那么对于每个连通分量，如果不存在奇数度的点，那么该分量是构成欧拉回路，只需要一个队伍就能走遍；如果奇数度的点有n个，那么需要n/2个队伍才能走遍，画一下图就很明了了。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; int g[MAXN]; int n, m; int f[MAXN]; int h[MAXN]; int r[MAXN]; vector<int> vs[MAXN]; void init() { clr(g); clr(h); rep(i,0,n) f[i] = i, vs[i].clear(); } int find(int x) { return f[x] == x ? x : (f[x] = find(f[x])); } void union_set(int a, int b) { int x = find(a); int y = find(b); f[x] = y; } int main() { int a, b; while(RII(n, m) != EOF) { init(); rep(i,1,m) { RII(a, b); g[a]++; g[b]++; union_set(a, b); } rep(i,1,n) find(i); int k = 0; rep(i,1,n) { if(!h[f[i]]) { h[f[i]] = ++k; r[k] = f[i]; vs[k].PB(i); } else vs[h[f[i]]].PB(i); } int cnt = 0; rep(i,1,k) { int tmp = 0; int num = vs[i].size(); rep(j,0,num-1) tmp += g[vs[i][j]]&1; if(tmp == 0 && num != 1) cnt++; else cnt += tmp / 2; } printf("%d ", cnt); } return 0; }
HDU 1116 Play on Words
对每个单词的首尾字母连边，看是否能构成欧拉通路或者欧拉回路，有两种做法：
1. 分别判断是否是欧拉通路和欧拉回路。
2. 如果可以构成欧拉回路，那答案是肯定的；否则设法将欧拉通路变为欧拉回路（找出欧拉通路首尾点，额外添加一条边构成欧拉回路），再计算一次，看是否能构成欧拉回路。
当然，两种做法都要先判断连通性，谨记欧拉回路是基于连通图的。
我用的第一种方法。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; int n, m, f[MAXN], g[MAXN]; char s[MAXM]; bool h[MAXN]; int v[MAXN]; // 记录出现的字母 int find(int x) { return f[x] == x ? x : (f[x] = find(f[x])); } void union_set(int u, int v) { int x = find(u); int y = find(v); f[x] = y; } int main() { int t, a, b; RI(t); while(t--) { RI(n); clr(g); clr(h); rep(i,0,26) f[i] = i; int k = 0; rep(i,1,n) { scanf("%s", s); a = s[0] - 'a'; b = s[strlen(s)-1] - 'a'; g[a]++; g[b]--; union_set(a, b); if(!h[a]) v[++k] = a, h[a] = true; if(!h[b]) v[++k] = b, h[b] = true; } bool flag = true; rep(i,1,k-1) if(find(v[i]) != find(v[i+1])) flag = false; int cnt1 = 0; // 记录出现度为1的节点的个数，即起点 int cnt2 = 0; // 记录出现度为-1的节点的个数，即终点 rep(i,0,25) { if(abs(g[i]) > 1) { flag = false; break; } if(g[i] == 1) cnt1++; if(g[i] == -1) cnt2++; } if(!flag || cnt1 > 1 || cnt2 > 1) printf("The door cannot be opened. "); else printf("Ordering is possible. "); } return 0; }
HDU 2894 DeBruijin
磁鼓欧拉回路。
举个具体例子吧，比如k=3的情况，设当前有一个结点为“010”，那么根据欧拉回路一笔画的想法，和该结点相连的结点应该是“100”和“101”，也就是原结点向左移一位然后最低位可以是“0”或“1”，根据这个思路，从“000”开始，先访问下一个低位为0的结点，在访问为1的结点，就能保证所得答案为最小，利用栈保存过程中遇到的可行结点，最后倒序输出栈即可。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; int n; bool vis[MAXN]; int stk[MAXN], top; void dfs(int u) { rep(i,0,1) { int t = ((u<<1)&((1<<n)-1)) + i; if(!vis[t]) { vis[t] = true; dfs(t); stk[top++] = i; } } } void init() { clr(vis); top = 0; } int main() { while(RI(n) != EOF) { init(); printf("%d ", 1 << n); dfs(0); rep(i,1,n-1) printf("0"); per(i,top-1,n-1) printf("%d", stk[i]); putchar(' '); } return 0; }
HDU 1956 Sightseeing tour
混合欧拉+最大流。
对于双向边，任意定向（至于为什么可以这样做，搞清楚求后面求最大流的过程后应该会很清楚了）。双向边用于建图（单向边不用于建图）。
然后是统计结点的度，计算差值 x = outg[i] - ing[i]，如果x>0，源点到 i 连一条边，权值为x/2，也就是说我们需要把与结点 i 相连的x/2条边反向，那么outg[i] == ing[i]；同理，如果x<0，连边 i 到汇点，权值为-x/2。最后，如果求出的最大流等于需要反向的总边数，也就是说该流网络满流，说明我们可以成功把想要反向的所有边都反向，那么就能符合题目要求了。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; struct Edge { int v, w, next; }E[MAXM]; int n, m; int head[MAXN], NE; int ing[MAXN], outg[MAXN]; int d[MAXN]; bool vis[MAXN]; int q[MAXN]; void add_edge (int u, int v, int w) { E[NE].v = v; E[NE].w = w; E[NE].next = head[u]; head[u] = NE++; E[NE].v = u; E[NE].w = 0; E[NE].next = head[v]; head[v] = NE++; } int BFS (int s, int t) { memset(d, 0x7f, sizeof(d)); memset(vis, false, sizeof(vis)); int qn = 0 ; d[s] = 0 ; vis[s] = true ; q[qn++] = s ; for (int qf = 0; qf < qn; ++qf ) { int u = q[qf]; for (int i = head[u]; i != -1; i = E[i].next) { int v = E[i].v; if (E[i].w > 0 && !vis[v]) { d[v] = d[u] + 1 ; vis[v] = true ; q[qn++] = v ; if (v == t) return d[t]; } } } return MAXN; } int DFS (int u, int df, int s, int t) { if (u == t) return df ; if (vis[u]) return 0 ; vis[u] = true; for (int i = head[u]; i != -1; i = E[i].next) { int v = E[i].v ; if (E[i].w > 0 && d[u] + 1 == d[v]) { int f = DFS (v, min(df, E[i].w), s , t); if ( f ) { E[i].w -= f ; E[i ^ 1].w += f ; return f ; } } } return 0 ; } int dinic (int s, int t) { int flow = 0 ; while (BFS (s, t) < MAXN) while ( true ) { clr(vis); int f = DFS (s, inf, s, t); if( !f ) break; flow += f ; } return flow; } void init() { clr(ing); clr(outg); NE = 0; memset(head, -1, sizeof head); } int main() { int a, b, c, t; RI(t); while(t--) { RII(n, m); init(); rep(i,1,m) { scanf("%d%d%d", &a, &b, &c); if(a == b) continue; outg[a]++; ing[b] ++; if(!c) add_edge(a, b, 1); } bool flag = true; rep(i,1,n) { int tot = ing[i] + outg[i]; if(tot & 1) flag = false; if(!flag) break; } if(!flag) { puts("impossible"); continue; } int sum = 0; rep(i,1,n) { int x = outg[i] - ing[i]; if(x > 0) { add_edge(0, i, x/2); sum += x/2; } if(x < 0) add_edge(i, n+1, -x/2); } if(dinic(0, n+1) == sum) puts("possible"); else puts("impossible"); } return 0; }
HDU 3472 HS BDC
此题和上一题类似，混和欧拉+最大流，不再赘述。、
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; struct Edge { int v, w, next; }E[MAXM]; int n, m; int head[MAXN], NE; char s[MAXN]; int ing[MAXN], outg[MAXN]; int f[MAXN]; vector<int> vs; bool h[MAXN]; int d[MAXN]; bool vis[MAXN]; int q[MAXN]; void add_edge (int u, int v, int w) { E[NE].v = v; E[NE].w = w; E[NE].next = head[u]; head[u] = NE++; E[NE].v = u; E[NE].w = 0; E[NE].next = head[v]; head[v] = NE++; } int BFS (int s, int t) { memset(d, 0x7f, sizeof(d)); memset(vis, false, sizeof(vis)); int qn = 0 ; d[s] = 0 ; vis[s] = true ; q[qn++] = s ; for (int qf = 0; qf < qn; ++qf ) { int u = q[qf]; for (int i = head[u]; i != -1; i = E[i].next) { int v = E[i].v; if (E[i].w > 0 && !vis[v]) { d[v] = d[u] + 1 ; vis[v] = true ; q[qn++] = v ; if (v == t) return d[t]; } } } return MAXN; } int DFS (int u, int df, int s, int t) { if (u == t) return df ; if (vis[u]) return 0 ; vis[u] = true; for (int i = head[u]; i != -1; i = E[i].next) { int v = E[i].v ; if (E[i].w > 0 && d[u] + 1 == d[v]) { int f = DFS (v, min(df, E[i].w), s , t); if ( f ) { E[i].w -= f ; E[i ^ 1].w += f ; return f ; } } } return 0 ; } int dinic (int s, int t) { int flow = 0 ; while (BFS (s, t) < MAXN) while ( true ) { clr(vis); int f = DFS (s, inf, s, t); if( !f ) break; flow += f ; } return flow; } int find(int x) { return f[x] == x ? x : (f[x] = find(f[x])); } void union_set(int u, int v) { int x = find(u); int y = find(v); f[x] = y; } void init() { clr(ing); clr(h); clr(outg); vs.clear(); NE = 0; memset(head, -1, sizeof head); rep(i,0,26) f[i] = i; } int main() { int t, a, b, c; RI(t); rep(cas,1,t) { RI(n); init(); rep(i,1,n) { scanf("%s%d", s, &c); a = s[0] - 'a' + 1; b = s[strlen(s) - 1] - 'a' + 1; if(!h[a]) vs.PB(a), h[a] = true; if(!h[b]) vs.PB(b), h[b] = true; if(a == b) continue; outg[a]++; ing[b] ++; union_set(a, b); if(c) add_edge(a, b, 1); } printf("Case %d: ", cas); bool flag = true; rep(i,0,vs.size()-2) { if(find(vs[i]) != find(vs[i+1])) flag = false; if(!flag) break; } if(!flag) { puts("Poor boy!"); continue; } int cnt = 0, a1, a2; rep(i,0,vs.size()-1) { int tot = ing[vs[i]] + outg[vs[i]]; if(tot & 1) { cnt++; if(cnt == 1) a1 = vs[i]; if(cnt == 2) a2 = vs[i]; } } if(cnt > 2) { puts("Poor boy!"); continue; } if(cnt == 2) { add_edge(a1, a2, 1); outg[a1]++; ing[a2]++; } int sum = 0; rep(i,0,vs.size()-1) { int x = outg[vs[i]] - ing[vs[i]]; if(x > 0) { add_edge(0, vs[i], x/2); sum += x/2; } if(x < 0) add_edge(vs[i], 27, -x/2); } if(dinic(0, 27) == sum) puts("Well done!"); else puts("Poor boy!"); } return 0; }
POJ 1041 John's trip
输出欧拉回路。
说白了就是一个dfs，代码很简单，不难理解。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; int x, y, z; int m[MAXN][MAXM]; int n, E, s, t; int g[MAXM]; bool vis[MAXM], tag; int stk[MAXM]; int top; void dfs(int u) { rep(i,1,E) { if(m[u][i] && !vis[i]) { vis[i] = true; dfs(m[u][i]); stk[top++] = i; } } } void init() { clr(g); clr(m); clr(vis); E = -1; s = inf; t = -1; top = 0; } int main() { while(RII(x, y) != EOF) { if(x == 0 && y == 0) break; RI(z); init(); E = max(E, z); s = min(s, min(x, y)); t = max(t, max(x, y)); m[x][z] = y; m[y][z] = x; g[x]++; g[y]++; while(RII(x, y) != EOF) { if(x == 0 && y == 0) break; RI(z); m[x][z] = y; m[y][z] = x; E = max(E, z); s = min(s, min(x, y)); t = max(t, max(x, y)); g[x]++; g[y]++; } bool flag = false; rep(i,1,t) { flag = g[i] & 1; if(flag) break; } if(flag) { puts("Round trip does not exist."); continue; } dfs(s); per(i,top-1,0) printf("%d%c", stk[i], i == 0 ? ' ' : ' '); } return 0; }
POJ 2230 Watchcow
还是输出欧拉回路，不过和POJ 1041有一点儿不同就是输出的是点，代码稍微有一点儿不同，需仔细体会。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; struct Node { int v, next; }E[MAXM]; int n, m; int NE, head[MAXN]; bool vis[MAXM]; void add_edge(int u, int v) { E[NE].v = v; E[NE].next = head[u]; head[u] = NE++; } void dfs(int u) { for(int i = head[u]; i != -1; i = E[i].next) { int v = E[i].v; if(!vis[i]) { vis[i] = true; dfs(v); } } printf("%d ", u); } void init() { NE = 0; clr(vis); memset(head, -1, sizeof head); } int main() { int a, b; RII(n, m); init(); rep(i,1,m) { RII(a, b); add_edge(a, b); add_edge(b, a); } dfs(1); return 0; }
POJ 2513 Colored Sticks
字典树+欧拉路径。
一开始用map存单词，爆了内存。后来才想起用字典树来标记单词，然后判断一下能否构成欧拉回路或欧拉通路即可。（以下代码采用泽阳师兄的字典树写法，比我原来的指针型动态分配内存写法快了一个数量级，此字典树写法才是标准的）。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; int g[MAXN]; int n, m, k; int f[MAXN]; struct Tree { int v; int next[26]; }P[1000005], root; int find(int x) { return f[x] == x ? x : (f[x] = find(f[x])); } void union_set(int u, int v) { int x = find(u); int y = find(v); f[x] = y; } int trie(char *s) { int h = 0; for(int i = 0; s[i]; i++) { int a = s[i] - 'a'; if(!P[h].next[a]) P[h].next[a] = ++k; h = P[h].next[a]; } if(P[h].v == 0) P[h].v = ++n; return P[h].v; } int main() { int a, b; char u[12], v[12]; clr(g); n = 0; k = 0; rep(i,0,MAXN) f[i] = i; while(scanf("%s%s", u, v) != EOF) { a = trie(u); b = trie(v); if(a == b) continue; g[a]++; g[b]++; union_set(a, b); } bool flag = true; rep(i,1,n-1) { if(find(i) != find(i+1)) { flag = false; break; } } if(!flag) { puts("Impossible"); return 0; } int cnt = 0; rep(i,1,n) cnt += g[i] & 1; if(cnt > 2) puts("Impossible"); else puts("Possible"); return 0; }
POJ 2337 Catenyms
因为要保证字典序最小，所以要先从大到小排一下序（没写错，是从大到小，因为到深搜递归的时候越小的字符串保存离栈顶越近），然后依然是打印出欧拉回路。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; struct Edge { int v, next; }E[MAXM]; int n, m; int head[MAXN], NE; string s[MAXM]; int ing[MAXN], outg[MAXN]; int f[MAXN]; vector<int> vs; int h[MAXM]; int stk[MAXM], top; bool vis[MAXM]; void add_edge (int u, int v) { E[NE].v = v; E[NE].next = head[u]; head[u] = NE++; } int find(int x) { return f[x] == x ? x : (f[x] = find(f[x])); } void union_set(int u, int v) { int x = find(u); int y = find(v); f[x] = y; } void init() { clr(h); clr(ing); clr(vis); clr(outg); vs.clear(); NE = 1; top = 0; memset(head, -1, sizeof head); rep(i,0,26) f[i] = i; } void dfs(int u, int d) { for(int i = head[u]; i != -1; i = E[i].next) { int v = E[i].v; if(!vis[i]) { vis[i] = true; dfs(v, i); } } if(d != -1) stk[top++] = d; } bool cmp(string u, string v) { return u >= v; } int check() { int num=0,cnt1=0,cnt2=0,id=-1; rep(j,0,vs.size()-1) { int i = vs[j]; if(find(i) == i) num++; if(ing[i]!=outg[i]) { if(ing[i]-outg[i]==1) cnt1++; else if(outg[i]-ing[i]==1) { cnt2++; id=i; } else return -1; } } if(num!=1) return -1; if(!((cnt1==1&&cnt2==1)||(cnt1==0&&cnt2==0))) return -1; if(id==-1) { for(int i=0;i<26;i++) { if(outg[i]>0) { id=i; break; } } } return id; } int main() { int t, a, b; cin >> t; while(t--) { RI(n); init(); rep(i,1,n) cin >> s[i]; sort(s + 1, s + 1 + n, cmp); rep(i,1,n) { a = s[i][0] - 'a' + 1; b = s[i][s[i].length() - 1] - 'a' + 1; if(!h[a]) vs.PB(a), h[a] = true; if(!h[b]) vs.PB(b), h[b] = true; outg[a]++; ing[b]++; union_set(a, b); add_edge(a, b); } int st = check(); if(st == -1) { cout << "***" << endl; continue; } dfs(st, -1); per(i,top-1,0) { cout << s[stk[i]]; if(i != 0) cout << '.'; } cout << endl; } return 0; }
POJ 1392 Ouroboros Snake
磁鼓欧拉回路，和前面HDU 2894类似。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 1<<20 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; int n, k, cnt; bool vis[MAXN]; int stk[MAXN], top; int ans[MAXN]; void dfs(int u) { rep(i,0,1) { int t = ((u<<1)&((1<<n)-1)) + i; if(!vis[t]) { vis[t] = true; dfs(t); stk[top++] = i; } } } void init() { clr(vis); top = 0; cnt = 0; } int main() { while(RII(n, k) != EOF) { if(n == 0 && k == 0) break; init(); dfs(0); rep(i,1,n-1) ans[cnt++] = 0; per(i,top-1,0) ans[cnt++] = stk[i]; int sum = 0; rep(i,k,k+n-1) sum = (sum<<1) | ans[i]; printf("%d ", sum); } return 0; }
POJ 1780 Code
这道题很蛋痛啊，要写非递归的磁鼓欧拉回路，不然要RE的。
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <algorithm> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <cmath> using namespace std; //#pragma comment(linker, "/STACK:102400000,102400000") #define pii pair<int,int> #define clr(a) memset((a),0,sizeof (a)) #define rep(i,a,b) for(int i=(a);i<=(int)(b);i++) #define per(i,a,b) for(int i=(a);i>=(int)(b);i--) #define inf 0x3f3f3f3f #define eps 1e-6 #define MAXN 10000007 #define MAXM 100000007 #define MOD 1000000007 #define debug puts("reach here") #define MP make_pair #define PB push_back #define RI(x) scanf("%d",&x) #define RII(x,y) scanf("%d%d",&x,&y) #define RIII(x,y,z) scanf("%d%d%d",&x,&y,&z) typedef long long LL; int n, k, cnt, p; int stk[MAXN], top; int ans[MAXN]; int num[MAXN]; void solve() { stk[top++] = 0; num[0]++; while(top) { int u = stk[--top]; ans[cnt++] = u % 10; u /= 10; while(num[u] < 10) { int w = u * 10 + num[u]; num[u]++; stk[top++] = w; u = w % p; } } } void init() { top = 0; cnt = 0; p = 1; rep(i,1,n-1) p *= 10; clr(num); } int main() { while(RI(n) != EOF) { if(n == 0) break; if(n == 1) {puts("0123456789"); continue;} init(); solve(); rep(i,1,n) printf("0"); per(i,cnt-1,n-1) printf("%d", ans[i]); rep(i,1,n-2) printf("0"); putchar(' '); } return 0; }