Problem One-Way Reform
题目大意
给一张n个点，m条边的无向图，要求给每条边定一个方向，使得最多的点入度等于出度，要求输出方案。
解题分析
最多点的数量就是入度为偶数的点。
将入度为奇数的点每两个组成一队，连一条无向边，之后求出欧拉回路即可。
参考程序
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <string> 8 #include <vector> 9 #include <cstdio> 10 #include <cstdlib> 11 #include <cstring> 12 #include <cassert> 13 #include <iostream> 14 #include <algorithm> 15 #pragma comment(linker,"/STACK:102400000,102400000") 16 using namespace std; 17 18 #define N 1000 19 #define E 50000 20 #define LL long long 21 #define lson l,m,rt<<1 22 #define rson m+1,r,rt<<1|1 23 #define clr(x,v) memset(x,v,sizeof(x)); 24 #define bitcnt(x) __builtin_popcount(x) 25 #define rep(x,y,z) for (int x=y;x<=z;x++) 26 #define repd(x,y,z) for (int x=y;x>=z;x--) 27 const int mo = 1000000007; 28 const int inf = 0x3f3f3f3f; 29 const int INF = 2000000000; 30 /**************************************************************************/ 31 32 int T,n,m,sum; 33 int lt[N],deg[N],f[N],dict[E]; 34 struct line{ 35 int u,v,nt,flag; 36 }eg[E]; 37 void add(int u,int v) 38 { 39 eg[++sum]=(line){u,v,lt[u],0}; lt[u]=sum; deg[v]++; 40 } 41 vector <int> vct,path; 42 stack <int> Q; 43 void dfs(int u) 44 { 45 int v=0; 46 Q.push(u); 47 f[u]=1; 48 for (int i=lt[u];i;i=eg[i].nt) 49 { 50 if (eg[i].flag) continue; 51 eg[i].flag=eg[i^1].flag=1; dict[i/2]=i; lt[u]=i; 52 v=eg[i].v; 53 dfs(v); 54 break; 55 } 56 } 57 void work(int S) 58 { 59 while (!Q.empty()) Q.pop(); 60 Q.push(S); 61 while (!Q.empty()) 62 { 63 int u=Q.top(),flag=0; Q.pop(); 64 for (int i=lt[u];i;i=eg[i].nt) 65 { 66 if (eg[i].flag) continue; 67 flag=1; 68 break; 69 } 70 if (flag) dfs(u); else path.push_back(u); 71 } 72 } 73 int main() 74 { 75 cin>>T; 76 while (T--) 77 { 78 cin>>n>>m; 79 int ans=n; 80 rep(i,1,n) deg[i]=lt[i]=0; sum=1; 81 rep(i,1,m) 82 { 83 int u,v; 84 scanf("%d%d",&u,&v); 85 add(u,v); add(v,u); 86 } 87 vct.clear(); path.clear(); 88 rep(i,1,n) if (deg[i] & 1) 89 { 90 ans--; 91 vct.push_back(i); 92 } 93 for (int i=0;i<vct.size();i+=2) 94 { 95 add(vct[i],vct[i+1]); 96 add(vct[i+1],vct[i]); 97 } 98 clr(f,0); 99 rep(i,1,n) if (f[i]==0) work(i); 100 printf("%d ",ans); 101 rep(i,1,m) printf("%d %d ",eg[dict[i]].u,eg[dict[i]].v); 102 //for (auto v:path) printf("%d ",v); 103 } 104 }