HDU5971 Wrestling Match

Wrestling Match

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 498    Accepted Submission(s): 226

Problem Description

Nowadays, at least one wrestling match is held every year in our country. There are a lot of people in the game is "good player”, the rest is "bad player”. Now, Xiao Ming is referee of the wrestling match and he has a list of the matches in his hand. At the same time, he knows some people are good players,some are bad players. He believes that every game is a battle between the good and the bad player. Now he wants to know whether all the people can be divided into "good player" and "bad player".

 

Input

Input contains multiple sets of data.For each set of data,there are four numbers in the first line:N (1 ≤ N≤ 1000)、M(1 ≤M ≤ 10000)、X,Y(X+Y≤N ),in order to show the number of players(numbered 1toN ),the number of matches,the number of known "good players" and the number of known "bad players".In the next M lines,Each line has two numbersa, b(a≠b) ,said there is a game between a and b .The next line has X different numbers.Each number is known as a "good player" number.The last line contains Y different numbers.Each number represents a known "bad player" number.Data guarantees there will not be a player number is a good player and also a bad player.

 

Output

If all the people can be divided into "good players" and "bad players”, output "YES", otherwise output "NO".

 

Sample Input

  

   5 4 0 0
1 3
1 4
3 5
4 5
5 4 1 0
1 3
1 4
3 5
4 5
2
  

 

Sample Output

  

   NO
YES
  

 

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）  

http://acm.split.hdu.edu.cn/showproblem.php?pid=5971

题目意思有点蛋痛 输入N个点 M对关系 后面是已经确定的好人个数  和坏人个数

每对关系里都有一个好人和一个坏人  在下面是已经确定的好人 和 坏人

问是否矛盾 (有点比较蛋痛 就是 a+b==0 OR N=1 的时候就直接NO )

(有的讲是否可以形成二分图) 我的做法是染色法  

跟下面这道题是一样的 不过这道题多了个条件就是把一些点确定了

所以从这些点开始染

http://blog.youkuaiyun.com/xky140610205/article/details/52763988

#include<bits/stdc++.h>
using namespace std;

vector<int>ap[1010];
int vis[1010],viss[1010];
int aa[1010],bb[1010];
int flagg;

int dfs(int u)
{
    int i;
    for(i=0; i<ap[u].size(); i++)
    {
        int v=ap[u][i];
        if(vis[v]==vis[u])
        {
            flagg=1;
            return 1;
        }
        if(!vis[v])
        {
            viss[v]=1;
            vis[v]=-vis[u];
            dfs(v);
        }
        if(flagg)
            return 1;
    }
    return 0;
}

int main()
{
    int n,m,a,b,i,c,d,flag;
    while(scanf("%d%d%d%d",&n,&m,&a,&b)!=EOF)
    {
        if(n==1)
        {
            printf("NO\n");
            continue;
        }
        memset(vis,0,sizeof(vis));
        memset(viss,0,sizeof(viss));
        for(i=0; i<1010; i++)
            ap[i].clear();
        while(m--)
        {
            scanf("%d%d",&c,&d);
            ap[c].push_back(d);
            ap[d].push_back(c);
        }
        for(i=0; i<a; i++)
        {
            scanf("%d",&c);
            aa[i]=c;
            vis[c]=1;
        }
        for(i=0; i<b; i++)
        {
            scanf("%d",&d);
            bb[i]=d;
            vis[d]=-1;
        }
        if(a+b==0)
        {
            printf("NO\n");
            continue;
        }

        flag=0;
        flagg=0;
        for(i=0; i<a; i++)
        {
            if(!viss[aa[i]])
            {
                flag=dfs(aa[i]);
                viss[aa[i]]=1;
            }
            if(flag)
                break;
        }

        if(!flag)
            for(i=0; i<b; i++)
            {
                if(!viss[bb[i]])
                {
                    flag=dfs(bb[i]);
                    viss[bb[i]]=1;
                }
                if(flag)
                    break;
            }

        for(i=1; i<=n; i++)
        {
            if(!viss[i])
            {
                viss[i]=1;
                vis[i]=1;
                flag=dfs(i);
            }
            if(flag)
                break;
        }
        if(flag)
            printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}