HDU2255 奔小康赚大钱 最大权值匹配

题意：此乃第一道真正意义上的最大权值匹配，其他题目其实都是求一个最小权值匹配。

代码如下：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

const int INF = 0x3f3f3f3f;

int N;
int w[305][305];
int lx[305], ly[305];
int sx[305], sy[305];
int match[305], slack[305];

int path(int u) {
    sx[u] = 1;
    for (int i = 1; i <= N; ++i) {
        if (sy[i]) continue;
        int t = lx[u] + ly[i] - w[u][i];
        if (!t) {
            sy[i] = 1;
            if (!match[i] || path(match[i])) {
                match[i] = u;
                return true;
            }
        } else {
            slack[i] = min(slack[i], t);
        }
    }
    return false;
}

void KM() {
    memset(match, 0, sizeof (match));
    memset(ly, 0, sizeof (ly));
    memset(lx, 0, sizeof (lx));
    for (int i = 1; i <= N; ++i) {
        for (int j = 1; j <= N; ++j) {
            lx[i] = max(lx[i], w[i][j]);    
        }
    }
    for (int i = 1; i <= N; ++i) {
        memset(slack, 0x3f, sizeof (slack));
        while (1) {
            memset(sx, 0, sizeof (sx));
            memset(sy, 0, sizeof (sy));
            if (path(i)) break;
            int d = INF;
            for (int j = 1; j <= N; ++j) {
                if (!sy[j]) d = min(d, slack[j]);
            }
            for (int j = 1; j <= N; ++j) {
                if (sx[j]) lx[j] -= d;
                if (sy[j]) ly[j] += d;
                else slack[j] -= d;
            }
        }
    }
    int ret = 0;
    for (int i = 1; i <= N; ++i) {
        ret += w[match[i]][i];    
    }
    printf("%d\n", ret);
}

int main() {
    while (scanf("%d", &N) != EOF) {
        for (int i = 1; i <= N; ++i) {
            for (int j = 1; j <= N; ++j) {
                scanf("%d", &w[i][j]);
            }
        }
        KM();
    }
    return 0;    
}