hdu4751 二分图的判断

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http://acm.hdu.edu.cn/showproblem.php?pid=4751

  This year is the 60th anniversary of NJUST, and to make the celebration more colorful, Tom200 is going to invite distinguished alumnus back to visit and take photos.
  After carefully planning, Tom200 announced his activity plan, one that contains two characters:
  1. Whether the effect of the event are good or bad has nothing to do with the number of people join in.
  2. The more people joining in one activity know each other, the more interesting the activity will be. Therefore, the best state is that, everyone knows each other.
  The event appeals to a great number of alumnus, and Tom200 finds that they may not know each other or may just unilaterally recognize others. To improve the activities effects, Tom200 has to divide all those who signed up into groups to take part in the activity at different time. As we know, one‘s energy is limited, and Tom200 can hold activity twice. Tom200 already knows the relationship of each two person, but he cannot divide them because the number is too large.
  Now Tom200 turns to you for help. Given the information, can you tell if it is possible to complete the dividing mission to make the two activity in best state.

3
3 0
1 0
1 2 0

YES

/**
hdu 4751 染色法二分图的判断
题目大意：给定一个些人物关系问是否能把所有的人分成两组，使得每一组的人都相互认识
解题思路：任意两个人如果不是相互认识就建边，然后采用染色的方法判断是否为二分图即可
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=105;

struct note
{
    int v,next;
}edge[maxn*maxn];

int head[maxn],ip;
int color[maxn];

void init()
{
    memset(head,-1,sizeof(head));
    ip=0;
}

void addedge(int u,int v)
{
    edge[ip].v=v,edge[ip].next=head[u],head[u]=ip++;
}

bool dfs(int u,int col)
{
    color[u]=col;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(color[v]!=-1)
        {
            if(color[v]==col)return false;
            continue;
        }
        if(!dfs(v,!col))return false;
    }
    return true;
}

int n;
int g[105][105];

int main()
{
    while(~scanf("%d",&n))
    {
        memset(g,0,sizeof(g));
        for(int i=1;i<=n;i++)
        {
            int x;
            while(~scanf("%d",&x))
            {
                if(x==0) break;
                g[i][x]=1;
            }
        }
        init();
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                if(g[i][j]==0||g[j][i]==0)
                {
                    addedge(i,j);
                    addedge(j,i);
                }
            }
        }
        memset(color,-1,sizeof(color));
        bool flag=true;
        for(int i=1;i<=n;i++)
        {
            if(color[i]==-1&&dfs(i,0)==false)
            {
                flag=false;
                break;
            }
        }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

hdu4751 二分图的判断

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原文地址：http://blog.csdn.net/lvshubao1314/article/details/44158463