染色法判定二分图
AcWing 860. 染色法判定二分图

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

using namespace std;

const int N = 100010, M = 200010;//无向图，边要多一倍

int n, m;
int h[N], e[M], ne[M], 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;  //把1变成2，把2变成1，就用3-c
        }
        else if (color[j] == c) return false;
    }

    return true;   
}

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))  //1号颜色
            {
                flag = false;
                break;
            }
        }

    if (flag) puts("Yes");
    else puts("No");

    return 0;
}

AcWing 861. 二分图的最大匹配

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

using namespace std;

const int N = 510, M = 100010;

int n1, n2, m;
//int h[N], e[N], ne[N], idx;//数组越界，什么错误都可能发生
int h[N], e[M], ne[M], 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; i = 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);//虽然是无向图，但是匹配找的时候，只会单向的去找
    }

    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;
}