Wrestling Match---hdu5971(2016CCPC大连 染色法判断是否是二分图)

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;
}