分配问题(二部图的最佳匹配 KM) 线性规划与网络流24题

//http://www.cnblogs.com/IMGavin/
#include <iostream>
#include <stdio.h>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <bitset>
#include <algorithm>
using namespace std;

typedef long long LL;
#define gets(A) fgets(A, 1e8, stdin)
const int INF = 0x3F3F3F3F, N = 108, MOD = 1003;

const double EPS = 1e-6;
int val[N][N];
bool visx[N], visy[N];
//始终有lx[x] + ly[y] >= w[x][y]，且结束后顶标之和最小
int lx[N], ly[N], slack[N], match[N];//match[y]表示Y集节点y对X的匹配
int nx, ny;//X集Y集节点个数，要先初始化

bool Hungary(int u){
	visx[u] = true;
	for(int i = 0; i < ny; i++){
		if(!visy[i]){
            if(lx[u] + ly[i] == val[u][i]){
                visy[i] = true;
                if(match[i] == -1 || Hungary(match[i])){
                    match[i] = u;
                    return true;
                }

            }else{
                slack[i] = min(slack[i], lx[u] + ly[i] - val[u][i]);
            }
		}
	}
	return false;
}

void KM(){
    memset(match, -1, sizeof(match));
	memset(lx, 0, sizeof(lx)); //初始化顶标
	memset(ly, 0, sizeof(ly)); //ly[i]为0
	 //lx[i]为权值最大的边
	for(int i = 0; i < nx; i++){
		for(int j = 0; j < ny; j++){
            lx[i] = max(lx[i], val[i][j]);
		}
	}
	for(int i = 0; i < nx; i++){
		for(int j = 0; j < ny; j++){
            slack[j] = INF;
		}
		while(1){
			memset(visx, 0, sizeof(visx));
			memset(visy, 0, sizeof(visy));
			if(Hungary(i)){
                break;
            }else{
				int tp = INF;
				for(int j = 0; j < ny; j++){
                    if(!visy[j] && tp > slack[j]){
                        tp = slack[j];
                    }
                }
				for(int j = 0; j < nx; j++){
					if(visx[j]){
                        lx[j] -= tp;
					}
				}

				for(int j = 0; j < ny; j++){
					if(visy[j]){
                        ly[j] += tp;
					}else{
                        slack[j] -= tp;
					}
				}
			}
		}
	}
}

int main(){
	int n;
	while(cin >> n){
		nx = ny = n;
		for(int i = 0; i < n; i++){
			for(int j = 0; j < n; j++){
				scanf("%d", &val[i][j]);
				val[i][j] = -val[i][j];
			}
		}
		KM();
		int ans = 0;
		for(int i = 0; i < n; i++){
			ans += val[match[i]][i];
		}

		cout<<-ans<<endl;

		for(int i = 0; i < n; i++){
			for(int j = 0; j < n; j++){
				val[i][j] = -val[i][j];
			}
		}
		KM();
		ans = 0;
		for(int i = 0; i < n; i++){
			ans += val[match[i]][i];
		}

		cout<<ans<<endl;
	}
	

	return 0;
}