POJ 2492 A Bug's Life【二分图染色法】

A Bug's Life

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 32828		Accepted: 10751

Description

Background 
Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs. 
Problem 
Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.

Input

The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.

Output

The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.

Sample Input

2
3 3
1 2
2 3
1 3
4 2
1 2
3 4

Sample Output

Scenario #1:
Suspicious bugs found!

Scenario #2:
No suspicious bugs found!

Hint

Huge input,scanf is recommended.

题目大意：有n只虫子，m个关系，关系表示x，y有性关系，如果有同性恋，输出 found，否则输出no found、

思路：m个关系表示是假设x,y这两个人是异性，如果有冲突，表示有同性恋，其实也就是说，不满足二分图。如果有不冲突，那么一个集合全是男的 ，一个集合全是女的，只有在两个集合之间有性关系，集合内是不应该有关系的，所以我们这里采用二分图染色的方法来搞定。

简单点说二分法就是对于这样的样例：

1 2

2 3

假设1是黑色，2是红色，3也是黑色，那么13同性 ，2与13异性，没有同性恋。

1 2

2 3

3 1

假设1是和黑色，那么2就是红色，通过2是红色，我们规定3是黑色，遍历3的时候，3和1同色，冲突，不满足二分图，即有同性恋。

AC代码：

#include<stdio.h>
#include<string.h>
#include<vector>
#include<queue>
using namespace std;
int f[1000005];
int color[1000005];
vector<int >map[1000005];
int bfs(int x)
{
    queue<int >s;
    s.push(x);
    if(color[x]==0)//这里也要注意，如果人家原先有颜色，千万不要覆盖他的颜色。
    color[x]=1;
    while(!s.empty())
    {
        int u=s.front();
        s.pop();
        for(int i=0;i<map[u].size();i++)
        {
            int v=map[u][i];
            if(color[v])//如果v有颜色
            {
                if(color[v]==color[u])//判断是否冲突
                {
                    return 0;
                }
            }
            else//如果v没有颜色
            {
                color[v]=3-color[u];//标记上于u相反的颜色
                s.push(v);
            }
        }
    }
    return 1;
}
int main()
{
    int t;
    int kase=0;
    scanf("%d",&t);
    while(t--)
    {
        memset(color,0,sizeof(color));
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            f[i]=i;
            map[i].clear();
        }
        for(int i=0;i<m;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            map[x].push_back(y);//首先这是个无向图，不要缺少条件。
            map[y].push_back(x);
        }
        int flag=0;
        for(int i=1;i<=n;i++)//遍历所有人、
        {
            if(bfs(i)==0)
            {
                flag=1;
                break;
            }
        }
        printf("Scenario #%d:\n",++kase);
        if(flag==1)
        {
            printf("Suspicious bugs found!\n");
        }
        else
        {
            printf("No suspicious bugs found!\n");
        }
        puts("");
    }
}