标签：list amp pac tor else 二分 others any out
InputFor each data set:
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.
Proceed to the end of file.
OutputIf these students cannot be divided into two groups, print "No". Otherwise, print the maximum number of pairs that can be arranged in those rooms.
Sample Input
4 4 1 2 1 3 1 4 2 3 6 5 1 2 1 3 1 4 2 5 3 6
Sample Output
No 3
题意：给n个数配对，分成两队，相同对的人不能认识；
题解：直接匈牙利算法，用二重循环判断是否有相同队又认识的人（居然没超时，我看网上都是bfs做的，感觉二重循环代码量少了很多）
#include<map> #include<set> #include<list> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define pi acos(-1) #define ll long long #define mod 1000000007 using namespace std; const double eps=1e-8; const int N=205,maxn=305,inf=0x3f3f3f3f; int n,m,color[N]; bool used[N],ok[N][N]; bool match(int x) { for(int i=1;i<=n;i++) { if(!used[i]&&ok[x][i]) { used[i]=1; if(color[i]==0||match(color[i])) { color[i]=x; return 1; } } } return 0; } int main() { while(cin>>n>>m){ memset(ok,0,sizeof(ok)); while(m--){ int a,b; cin>>a>>b; ok[a][b]=ok[b][a]=1; } memset(color,0,sizeof(color)); int num=0; for(int i=1;i<=n;i++) { memset(used,0,sizeof(used)); if(match(i))num++; } // for(int i=1;i<=n;i++)cout<<color[i]<<" "; bool flag=0; for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++) if(ok[i][color[j]]&&ok[i][j]) flag=1; if(!flag)cout<<num/2<<endl; else cout<<"No"<<endl; } return 0; }
标签：list amp pac tor else 二分 others any out
原文地址：http://www.cnblogs.com/acjiumeng/p/6731093.html
