一、如何判断一个图是二分图？

二、代码

1、判断是否为二分图

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 200010;

int n, m;
int h[N], e[N], ne[N], idx;
int color[N];

void add(int a, int b)
{
	e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

bool dfs(int u, int c)
{
	color[u] = c;
	for(int i = h[u]; i != -1; i = ne[i]){
		int j = e[i];
		if(!color[j]){
			if(!dfs(j, 3-c))
				return false;
		}
		else if(color[j] == c)
			return false;
	}
}

int main()
{
	scanf("%d%d", &n, &m);

	memset(h, -1, sizeof h);

	while(m--){
		int a, b;
		scanf("%d%d", &a, &b);
		add(a, b), add(b, a);
	}

	bool flag = true;
	for(int i = 1; i <= n; i++)
		if(!color[i]){
			if(!dfs(i, 1)){
				flag = false;
				break;
			}
		}


	if(flag)
		puts("Yes");
	else
		puts("No");
	
	return 0;
}

2、匈牙利算法

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 510, M = 100010;

int n1, n2, m;
int h[N], e[N], ne[N], idx;
int match[N];
bool st[N];

void add(int a, int b)
{
	e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

bool find(int x)
{
	for(int i = h[x]; i != -1; ne[i]){
		int j = e[i];
		if(!st[j]){
			st[j] = true;
			if(match[j] == 0 || find(match[j])){
				match[j] = x;
				return true;
			}
		}
	}

	return false;
}

int main()
{
	scanf("%d%d%d", &n1, &n2, &m);

	memset(h, -1, sizeof h);

	while(m--){
		int a, b;
		scanf("%d%d", &a, &b);
		add(a, b);
	}

	int res = 0;
	for(int i = 1; i <= n1; i++){
		memset(st, false, sizeof st);
		if(find(i))
			res ++;
	}

	printf("%d\n", res);
	
	return 0;
}