HDU 5971 icpc 大连 A Wrestling Match

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Wrestling Match

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 68 Accepted Submission(s): 37

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

Sample Output

NO
YES

Source

题目大意：

给出 n 个人，m 场比赛 x 个已经确定的好人 y 个已经确定的坏人。

每场比赛由好人和坏人组成。

问是否能够将每个人划分成好人或者坏人。

思路：

dfs 染色。先从已经确定的人开始染色与其相关的人。矛盾就 NO

然后对于不确定的人，枚举染色。如果矛盾就是 NO

如果最后还存在不确定的，那么就是 NO

否则就是 YES

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

const int MAX_V = 1005;

vector<int> G[MAX_V];
int V;
int color[MAX_V];

bool dfs(int v, int c)
{
    color[v] = c;
    for (int i = 0; i < G[v].size(); ++i)
    {
        if (color[G[v][i]] == c)
        {
            return false;
        }
        if (color[G[v][i]] == 0 && !dfs(G[v][i], -c))
            return false;
    }
    return true;
}

void solve()
{
    for (int i = 0; i < V; ++i)
    {
        if(color[i]==1)
        {
            if(!dfs(i,1))
            {
                puts("NO");
                return ;
            }
        }
        else if(color[i]==-1)
        {
            if(!dfs(i,-1))
            {
                puts("NO");
                return ;
            }
        }
    }

    for (int i = 0; i < V; ++i)
    {
        if(color[i]==0)
        {
            if(!dfs(i,1))
            {
                puts("NO");
                return ;
            }
        }
    }
    for(int i=0;i<V;i++)
    {
        if(color[i]==-5)
        {

            puts("NO");
                return ;
        }

    }
    puts("YES");
}

int main()
{
    int E;
    int x,y;
    while (~scanf("%d%d%d%d", &V, &E,&x,&y))
    {
        int a, b;
        for (int i = 0; i < V; ++i)
        {
            G[i].clear();
            color[i]=-5;
        }
        for (int i = 0; i < E; ++i)
        {
            scanf("%d%d", &a, &b);
            a--;
            b--;
            color[a]=0;
            color[b]=0;
            G[a].push_back(b);
            G[b].push_back(a);
        }
        for(int i=0;i<x;i++)
        {
            scanf("%d",&a);
            a--;
            color[a]=1;
        }
        for(int i=0;i<y;i++)
        {
            scanf("%d",&a);
            a--;
            color[a]=-1;
        }
        solve();
    }
    return 0;
}