C - 小明系列故事――捉迷藏
这个题目看了一下题解，感觉没有很难，应该是可以自己敲出来的，感觉自己好蠢。。。
这个是一个bfs 用bfs就很好写了，首先可以预处理出大明和二明能被发现的位置，标记一下。
然后跑bfs，注意这个bfs记录一下状态，记录一下是否看到了大明和二明。
这个题目和之前写的旅游这个题目很像，所以还是很好写的
#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <queue> #include <vector> #include <iostream> #include <string> #define inf 0x3f3f3f3f #define inf64 0x3f3f3f3f3f3f3f3f using namespace std; const int maxn = 1e5 + 10; typedef long long ll; int n, m, t; char s[110][110]; bool flaga[110][110], flagb[110][110]; void checka(int x,int y) { int i = x, j = y; while (s[i][j] != 'X'&&i >= 1 && s[i][j] != 'E') flaga[i][j] = 1, i--; i = x; while (s[i][j] != 'X'&&i <= n && s[i][j] != 'E') flaga[i][j] = 1, i++; i = x; while (s[i][j] != 'X'&&j <= m && s[i][j] != 'E') flaga[i][j] = 1, j++; j = y; while (s[i][j] != 'X'&&j >= 1 && s[i][j] != 'E') flaga[i][j] = 1, j--; } void checkb(int x,int y) { int i = x, j = y; while (s[i][j] != 'X'&&i >= 1 && s[i][j] != 'D') flagb[i][j] = 1, i--; i = x; while (s[i][j] != 'X'&&i <= n && s[i][j] != 'D') flagb[i][j] = 1, i++; i = x; while (s[i][j] != 'X'&&j <= m && s[i][j] != 'D') flagb[i][j] = 1, j++; j = y; while (s[i][j] != 'X'&&j >= 1 && s[i][j] != 'D') flagb[i][j] = 1, j--; } int dx[] = { 0,1,-1,0 }; int dy[] = { 1,0,0,-1 }; struct node { int x, y, now; node(int x=0,int y=0,int now=0):x(x),y(y),now(now){} }; int dp[110][110][5]; bool vis[110][110][5]; int bfs(int sx,int sy) { int ans = inf; queue<node>que; int tmp = 0; memset(vis, 0, sizeof(vis)); if (flaga[sx][sy]) tmp |= (1 << 0); if (flagb[sx][sy]) tmp |= (1 << 1); dp[sx][sy][tmp] = 0; que.push(node(sx, sy, tmp)); while(!que.empty()) { node u = que.front(); que.pop(); int x = u.x, y = u.y, now = u.now; if (now == 3)ans = min(ans, dp[x][y][now]); for(int i=0;i<4;i++) { int tx = x + dx[i]; int ty = y + dy[i]; int tmp1 = now; if (tx<1 || ty<1 || tx>n || ty>m) continue; if (s[tx][ty] == 'X') continue; if (s[tx][ty] == 'E') continue; if (s[tx][ty] == 'D') continue; if (flaga[tx][ty]) tmp1 |= (1 << 0); if (flagb[tx][ty]) tmp1 |= (1 << 1); if (vis[tx][ty][tmp1]) continue; vis[tx][ty][tmp1] = 1; dp[tx][ty][tmp1] = dp[x][y][now] + 1; que.push(node(tx, ty, tmp1)); } } return ans; } int main() { int tim; scanf("%d", &tim); for(int cas=1;cas<=tim;cas++) { int sx = 0, sy = 0; scanf("%d%d%d", &n, &m, &t); memset(flaga, 0, sizeof(flaga)); memset(flagb, 0, sizeof(flagb)); for (int i = 1; i <= n; i++) { scanf("%s", s[i] + 1); } for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (s[i][j] == 'D') checka(i, j); if (s[i][j] == 'E') checkb(i, j); if (s[i][j] == 'S') sx = i, sy = j; } } int ans=bfs(sx, sy); printf("Case %d: ", cas); if (ans <= t) printf("%d ", ans); else printf("-1 "); } return 0; }
这个题目大意是：
db爱好运动，但是单纯的运动会使得他很枯燥，现在他想边跑步边看风景。已知现在有n个风景点（编号为1号~n号），同时有m条道路将这n个风景点连接起来。
这些风景点总共有3类：A，B，C；为了方便表示，我们令 A=0，B=1，C=2。db一开始在1号风景点（可以为A，B，C类）。现在db想在跑步的过程中经过至少一个B类风景点的同时至少经过一个C类风景点，最后再回到1号风景点。现在db想要在尽可能短的时间内跑步结束，你能帮他找出一条路程最短同时满足题目条件的路径吗？
这个就是一个最短路的时候记录状态即可。
#include <algorithm> #include <cstdio> #include <iostream> #include <cstdlib> #include <string> #include <cstring> #include <queue> #include <vector> #define inf 0x3f3f3f3f #define inf64 0x3f3f3f3f3f3f3f3f using namespace std; const int maxn = 4e5 + 10, sum = (1 << 3); typedef long long ll; ll d[maxn][1 << 3]; int n, m, num[maxn]; bool vis[maxn][1 << 3]; struct edge { int from, to;ll dist; edge(int from=0, int to=0, ll dist=0) :from(from), to(to), dist(dist) {} }; struct heapnode { int u, tmp; ll d; heapnode(ll d = 0, int u = 0,int tmp = 0) : d(d), u(u), tmp(tmp) {} bool operator<(const heapnode &a) const { return a.d < d; } }; vector<edge> e; vector<int>G[maxn]; void add(int u,int v,ll w) { e.push_back(edge(u, v, w)); e.push_back(edge(v, u, w)); int len = e.size(); G[u].push_back(len - 2); G[v].push_back(len - 1); } ll ans = 0; void dijkstra(int s) { priority_queue<heapnode>que; for(int i=0;i<=n;i++) for (int j = 0; j < sum; j++) d[i][j] = inf64; memset(vis, 0, sizeof(vis)); que.push(heapnode(0, s, 1 << num[s])); d[s][1<<num[s]] = 0; while (!que.empty()) { heapnode x = que.top(); que.pop(); int u = x.u, tmp1 = x.tmp; if (vis[u][tmp1]) continue; vis[u][tmp1] = 1; if (tmp1 >= 6 && u == 1) ans = min(ans, x.d); for(int j=0;j<G[u].size();j++) { edge now = e[G[u][j]]; int v = now.to; int tmp = 1 << num[v]; tmp |= tmp1; if(d[v][tmp] >x.d + now.dist &&!vis[v][tmp]) { d[v][tmp] = x.d + now.dist; que.push(heapnode(d[v][tmp], v, tmp)); } } } } int main() { scanf("%d%d", &n, &m); for (int i = 1; i <= n; i++) scanf("%d", &num[i]); while(m--) { int u, v; ll w; scanf("%d%d%lld", &u, &v, &w); add(u, v, w); } ans = inf64; dijkstra(1); printf("%lld ", ans); return 0; }