《算法基础》 二分图
1.染色法判断二分图

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 100010, M = 200010;
int h[N], e[M], ne[M], idx;
int color[N];
int n, m;
void add(int a, int b)
{
e[idx] = b; ne[idx] = h[a]; h[a] = idx ++;
}
int 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;
}
return true;
}
int main()
{
cin >> 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 == true) cout << "Yes" << endl;
else cout << "No" << endl;
return 0;
}
2.匈牙利算法

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 510, M = 100010;
int h[N], e[M], ne[M], idx;
int match[N];
bool st[N];
int n1, n2, m;
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] = 1;
if (match[j] == 0 || find(match[j])) {
match[j] = x;
return true;
}
}
}
return false;
}
int main()
{
cin >> 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, 0, sizeof st);
if (find(i)) res ++;
}
cout << res << endl;
return 0;
}