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doubleconst 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;
}