题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5971
题意:有n个人,编号为1-n, 已知X个人是good,Y个人是bad,m场比赛,每场比赛都有一个good和一个bad人结合起来,问这n个人是否能被分成两种人
其实就是判断是否为二分图,用染色法判断一下就可以了
#include <stdio.h> #include <string.h> #include <math.h> #include <algorithm> #include <bitset> #include <iostream> #include <time.h> #include <vector> #include <queue> typedef long long LL; using namespace std; const int N = 10265; const double eps = 1e-10; const int INF = 0x3f3f3f3f; const int mod = 1000000007; const double PI = 4*atan(1.0); int a[N], b[N], f[N]; int main() { int n, m, X, Y, x; while(scanf("%d %d %d %d", &n, &m, &X, &Y) != EOF) { memset(f, -1, sizeof(f)); for(int i=1; i<=m; i++) scanf("%d %d", &a[i], &b[i]); for(int i=1; i<=X; i++) { scanf("%d", &x); f[x] = 1; } for(int i=1; i<=Y; i++) { scanf("%d", &x); f[x] = 0; } int flag = 0; for(int i=1; i<=m; i++) { if(f[a[i]] == -1 && f[b[i]] == -1) f[a[i]] = 1, f[b[i]] = 0; else if(f[a[i]] != -1 && f[b[i]] == -1) f[b[i]] = f[a[i]]^1; else if(f[a[i]] == -1 && f[b[i]] != -1) f[a[i]] = f[b[i]]^1; else if(f[a[i]] == f[b[i]] && f[a[i]]!=-1) flag = 1; } for(int i=1; i<=n; i++) { if(f[i] == -1) flag = 1; } if(flag) puts("NO"); else puts("YES"); } return 0; }