（二分最大权匹配）奔小康赚大钱-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_40588429/article/details/89319494

本文深入探讨了KM算法，一种解决稳定匹配问题的有效方法。通过详细的代码解释，展示了如何使用KM算法来解决最大收入配对问题，即为每个人找到最适合的房屋以最大化总收入。文章包含了完整的C++实现代码，对理解KM算法及其应用提供了清晰的指导。

http://acm.hdu.edu.cn/showproblem.php?pid=2255

题意给出n个人和n个房屋，给出每个人对每个人房屋的出价，求出最大收入
KM算法https://www.cnblogs.com/logosG/p/logos.html

#include <bits/stdc++.h>
using namespace std;
const int maxn = 300 + 5;
const int INF = 0x3f3f3f3f;
int n, nx, ny;
int linker[maxn], lx[maxn], ly[maxn], slack[maxn];
int visx[maxn], visy[maxn], w[maxn][maxn];

int dfs(int x) {
    visx[x] = 1;
    for(int y = 1; y <= ny; ++y) {
        if(visy[y]) continue;
        int tmp = lx[x] + ly[y] - w[x][y];
        if(tmp == 0) {
            visy[y] = 1;
            if(linker[y] == -1 || dfs(linker[y])) {
                linker[y] = x;
                return 1;
            }
        }
        else if(slack[y] > tmp) slack[y] = tmp;
    }
    return 0;
}
int KM() {
    int i, j;
    memset(linker, -1, sizeof(linker));
    memset(ly, 0, sizeof(ly));
    for(i = 1; i <= nx; ++i)
        for(j = 1, lx[i] = -INF; j <= ny; ++j)
            if(w[i][j] > lx[i]) lx[i] = w[i][j];
    for(int x = 1; x <= nx; ++x) {
        for(i = 1; i <= ny; ++i)
            slack[i] = INF;
        while(1) {
            memset(visx, 0, sizeof(visx));
            memset(visy, 0, sizeof(visy));
            if(dfs(x)) break;
            int d = INF;
            for(i = 1; i <= ny; ++i)
                if(!visy[i] && d > slack[i])
                    d = slack[i];
            for(i = 1; i <= nx; ++i)
                if(visx[i]) lx[i] -= d;
            for(i = 1; i <= ny; ++i)
                if(visy[i]) ly[i] += d;
                else slack[i] -= d;
        }
    }
    int res = 0;
    for(i = 1; i <= ny; ++i)
        if(linker[i] != -1)
            res += w[linker[i]][i];
    return res;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    while(cin >> n) {
        nx = ny = n;
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
                cin >> w[i][j];
        int ans = KM();
        cout << ans << endl;
    }
}