【模板】二分图最大权完美匹配（KM算法）/洛谷P6577

1|0题目链接

https://www.luogu.com.cn/problem/P6577

2|0题目大意

给定一个二分图，其左右点的个数各为 $n$ ，带权边数为 $m$ ，保证存在完美匹配。

求一种完美匹配的方案，使得最终匹配边的边权之和最大。

3|0题目解析

二分图最大权完美匹配，一般用 $K M$ 算法完成。

基础版的 $K M$ 算法，是匈牙利算法的改进，依然采用 $D F S$ 策略，速度较慢。

而改进版的 $K M$ 算法采用 $B F S$ 策略，有效提升速度。

点数为 $n$ ，边数为 $m$ 。

时间复杂度： $O (n^{2} m)$

4|0参考代码

基础版（会超时）

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 505;
int mp[N][N], lx[N], ly[N], visx[N], visy[N], match[N];
int n, m, minz, k;

bool dfs(int x, int K)
{
    visx[x] = K;
    for (int y = 1; y <= n; ++y) {
        if (visy[y] != K && mp[x][y] != INF) {
            int t = lx[x] + ly[y] - mp[x][y];
            if (!t) {
                visy[y] = K;
                if (!match[y] || dfs(match[y], K)) {
                    match[y] = x;
                    return true;
                }
            }
            else minz = min(minz, t);
        }
    }
    return false;
}
void KM()
{
    for (int i = 1; i <= n; ++i) {
        while (1) {
            minz = INF;
            if (dfs(i, ++k)) break;
            for (int j = 1; j <= n; ++j) {
                if (visx[j] == k) lx[j] -= minz;
                if (visy[j] == k) ly[j] += minz;
            }
        }
    }
}
int main()
{
    scanf("%d%d", &n, &m);
    memset(mp, INF, sizeof mp);
    for (int i = 1; i <= n; ++i) lx[i] = -INF;
    for (int i = 0; i < m; ++i) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        mp[u][v] = w;
        lx[u] = max(lx[u], w);
    }
    KM();
    ll ans = 0;
    for (int i = 1; i <= n; ++i) ans += mp[match[i]][i];
    printf("%lld\n", ans);
    for (int i = 1; i <= n; ++i) printf("%d ", match[i]);
    putchar('\n');
    return 0;
}

改进版

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 505;
int mp[N][N], lx[N], ly[N], visy[N], matchy[N], pr[N], slack[N];
int n, m, minz, k;

void bfs(int x, int K)
{
    int y = 0, yy = 0;
    memset(pr, 0, sizeof pr);
    memset(slack, INF, sizeof slack);
    matchy[y] = x;
    while (true)
    {
        x = matchy[y];
        visy[y] = K;
        for (int i = 1; i <= n; ++i)
        {
            if (visy[i] == K || mp[x][i] == -INF) continue;
            int t = lx[x] + ly[i] - mp[x][i];
            if (slack[i] > t)
            {
                slack[i] = t;
                pr[i] = y;
            }
        }
        minz = INF;
        for (int i = 1; i <= n; ++i) {
            if (visy[i] != K && slack[i] < minz) {
                minz = slack[i];
                yy = i;
                if (!minz) break;
            }
        }
        if (minz) {
            for(int i = 0; i <= n; ++i)
            {
                if (visy[i] == K) lx[matchy[i]] -= minz, ly[i] += minz;
                else slack[i] -= minz;
            }
        }
        y = yy;
        if (!matchy[y]) break;
    }
    while (y) {
        matchy[y] = matchy[pr[y]];
        y = pr[y];
    }
}
void KM()
{
    for (int i = 1; i <= n; ++i) {
        bfs(i, ++k);
    }
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            mp[i][j] = -INF;
        }
    }
    for (int i = 1; i <= n; ++i) lx[i] = -INF;
    for (int i = 0; i < m; ++i) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        mp[u][v] = w;
        lx[u] = max(lx[u], w);
    }
    KM();
    ll ans = 0;
    for (int i = 1; i <= n; ++i) ans += mp[matchy[i]][i];
    printf("%lld\n", ans);
    for (int i = 1; i <= n; ++i) printf("%d ", matchy[i]);
    putchar('\n');
    return 0;
}