二分图的最佳完美匹配（模板）

        二分图的最佳完美匹配，也就是带权值的无向二分图中权值之和最大的完美匹配，整个图分为两个不相交的集合x和y，采用KM算法求解，也称匈牙利算法. 时间复杂度为O(n^3)

typedef int type;//实际数据类型,一般是int或double

const type inf = 2e9;
const double eps = 1e-8;
const int maxn = 330;

int n, m;//x中结点个数，y中结点个数，下标从0开始
int matchx[maxn], matchy[maxn];//matchx[i]指的是xi匹配到的y中结点的编号,matchy[j]指的是yj匹配到的x中结点的编号
int visx[maxn], visy[maxn];//访问标记数组
type lx[maxn], ly[maxn];//x,y中每个点的期望值
type w[maxn][maxn];//类似邻接矩阵w[x][y]表示x到y有权值为w[x][y]的边
type slack[maxn];//y中每个元素的slack值

bool dfs(int x) {
    visx[x] = 1;
    for (int y = 0; y < m; ++y) {
        if (visy[y]) continue;
        type tmp = lx[x] + ly[y] - w[x][y];
        if (tmp == 0) {//double的话改成: if (fabs(tmp)<=eps)
            visy[y] = 1;
            if (matchy[y] == -1 || dfs(matchy[y])) {
                matchx[x] = y;
                matchy[y] = x;
                return true;
            }
        }
        else {
            slack[y] = min(slack[y], tmp);
        }
    }
    return false;
}

void KM() {
    memset(matchy, -1, sizeof(matchy));
    memset(ly, 0, sizeof(ly));
    for (int i = 0; i < n; ++i) {
        lx[i] = -inf;
        for (int j = 0; j < m; ++j) {
            lx[i] = max(lx[i], w[i][j]);
        }
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) slack[j] = inf;
        while (1) {
            memset(visx, 0, sizeof(visx));
            memset(visy, 0, sizeof(visy));
            if (dfs(i)) break;
            type d = inf;
            for (int j = 0; j < m; ++j) {
                if (!visy[j]) d = min(d, slack[j]);
            }
            for (int j = 0; j < n; ++j) { if (visx[j]) lx[j] -= d; }
            for (int j = 0; j < m; ++j) {
                if (visy[j]) ly[j] += d;
                else slack[j] -= d;
            }
        }
    }
}

        模板题 hdu2255 http://acm.hdu.edu.cn/showproblem.php?pid=2255

        中文题目，裸的二分图最佳完美匹配问题

#include<bits/stdc++.h>
using namespace std;

typedef int type;

const type inf = 2e9;
const double eps = 1e-8;
const int maxn = 330;

int n, m;
int matchx[maxn], matchy[maxn];
int visx[maxn], visy[maxn];
type lx[maxn], ly[maxn];
type w[maxn][maxn];
type slack[maxn];

bool dfs(int x) {
    visx[x] = 1;
    for (int y = 0; y < m; ++y) {
        if (visy[y]) continue;
        type tmp = lx[x] + ly[y] - w[x][y];
        if (tmp == 0) {
            visy[y] = 1;
            if (matchy[y] == -1 || dfs(matchy[y])) {
                matchx[x] = y;
                matchy[y] = x;
                return true;
            }
        }
        else {
            slack[y] = min(slack[y], tmp);
        }
    }
    return false;
}

void KM() {
    memset(matchy, -1, sizeof(matchy));
    memset(ly, 0, sizeof(ly));
    for (int i = 0; i < n; ++i) {
        lx[i] = -inf;
        for (int j = 0; j < m; ++j) {
            lx[i] = max(lx[i], w[i][j]);
        }
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) slack[j] = inf;
        while (1) {
            memset(visx, 0, sizeof(visx));
            memset(visy, 0, sizeof(visy));
            if (dfs(i)) break;
            type d = inf;
            for (int j = 0; j < m; ++j) {
                if (!visy[j]) d = min(d, slack[j]);
            }
            for (int j = 0; j < n; ++j) { if (visx[j]) lx[j] -= d; }
            for (int j = 0; j < m; ++j) {
                if (visy[j]) ly[j] += d;
                else slack[j] -= d;
            }
        }
    }
}

int main() {
    while (scanf("%d", &n) == 1) {
        m = n;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                scanf("%d", &w[i][j]);
            }
        }
        KM();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += w[i][matchx[i]];
        }
        printf("%d\n", ans);
    }
    return 0;
}