一次比赛的题，以前都是匈牙利算法处理二分图问题（即已知是二分图），这次是判断二分图，注意处理方式的选择。
Mediacy
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 103 Accepted Submission(s): 18
Problem Description
In yrleep`s class,there are only two kinds of relationship between the classmates which can not be passed,namely incompatible and live in harmony.So yrleep wants to put them into two parts to make both of them get along well with each other in each part . Can his ideas be implemented?
Input
The input consists of multiple test cases.the first line input two integers n,m,n indicates the number of people
The next m lines,each line has two integers a,b,indicates a and b are incompatible.
Output
If his idea can come true ,output yes.if not,output no.
Sample Input
5 3
1 2
3 4
4 5
Sample Output
yes
#include <cstdio>
#include <iostream>
#include <memory.h>
#include <vector>
using namespace std;
const int maxn = 10000 + 10;
vector G[maxn];
int n, m;
int color[maxn];
bool dfs(int v, int c)
{
color[v] = c;
for(int i=0; i< G[v].size(); i++) {
//如果相邻的顶点同色，则返回false
if(color[G[v][i]] == c) return false;
//如果相邻的顶点还没有被染色，则染成-c
if(color[G[v][i]] == 0 && !dfs(G[v][i], -c)) return false;
}
return true;
}
void solve()
{
for(int i=0; i
if(color[i] == 0) {
if(!dfs(i,1)) {
printf("no\n");
return ;
}
}
}
printf("yes\n");
}
int main()
{
// freopen("in.txt","r",stdin);
int i, x, y;
while(~scanf("%d%d",&n,&m)) {
memset(color, 0, sizeof (color) );
for(i=1; i<=n; ++i) G[i].clear();
for(i=1; i<=m; ++i) {
scanf("%d%d",&x,&y);
G[x].push_back(y);
G[y].push_back(x);
}
solve();
}
return 0;
}
原文：http://www.cnblogs.com/balfish/p/4014409.html