POJ 2400 Supervisor, Supervisee（KM二分图最大权值匹配）题解

题意：n个老板n个员工，先给你n*n的数据，i行j列代表第i个老板第j喜欢的员工是谁，再给你n*n的数据，i行j列代表第i个员工第j喜欢的老板是谁，如果匹配到第k喜欢的人就会产生一个分数k-1。现在让你给老板和员工配对，希望得到的分数的平均数最少，并给出哪个老板匹配哪个员工，多种情况按字典序输出。

思路：题目中的input提示是错的...

这题就是km最大权值匹配的裸题，分数最小那就把权值变负，然后跑出最少的总分。因为n比较小，可以dfs求出所有情况。

代码：

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 50 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int nx, ny;
int g[maxn][maxn];
int linker[maxn], lx[maxn], ly[maxn];
int slack[maxn];
bool visx[maxn], visy[maxn];
bool dfs(int x){
    visx[x] = true;
    for(int y = 0; y < ny; y++){
        if(visy[y]) continue;
        int tmp = lx[x] + ly[y] - g[x][y];
        if(tmp == 0){
            visy[y] = true;
            if(linker[y] == -1 || dfs(linker[y])){
                linker[y] = x;
                return true;
            }
        }
        else if(slack[y] > tmp){
            slack[y] = tmp;
        }
    }
    return false;
}
int km(){
    memset(linker, -1, sizeof(linker));
    memset(ly, 0, sizeof(ly));
    for(int i = 0; i < nx; i++){
        lx[i] = -INF;
        for(int j = 0; j < ny; j++){
            if(g[i][j] > lx[i]){
                lx[i] = g[i][j];
            }
        }
    }
    for(int x = 0; x < nx; x++){
        for(int i = 0; i < ny; i++)
            slack[i] = INF;
        while(true){
            memset(visx, false, sizeof(visx));
            memset(visy, false, sizeof(visy));
            if(dfs(x)) break;
            int d = INF;
            for(int i = 0; i < ny; i++)
                if(!visy[i] && d > slack[i])
                    d = slack[i];
            for(int i = 0; i < nx; i++)
                if(visx[i])
                    lx[i] -= d;
            for(int i = 0; i < ny; i++){
                if(visy[i]) ly[i] += d;
                else slack[i] -= d;
            }
        }
    }
    int res = 0;
    for(int i = 0; i < ny; i++){
        if(linker[i] != -1)
            res += g[linker[i]][i];
    }
    return res;
}
int ans[maxn], vis[maxn];
int n, t, ret, num, ca = 1;
void DFS(int u, int sco){
    if(sco < ret) return;
    if(u == n){
        if(sco == ret){
            printf("Best Pairing %d\n", num++);
            for(int i = 0; i < n; i++){
                printf("Supervisor %d with Employee %d\n", i + 1, ans[i]);
            }
        }
        return;
    }
    for(int i = 0; i < n; i++){
        if(vis[i]) continue;
        vis[i] = 1;
        ans[u] = i + 1;
        DFS(u + 1, sco + g[u][i]);
        vis[i] = 0;
    }
}
int main(){
    scanf("%d", &t);
    while(t--){
        num = 0;
        scanf("%d", &n);
        memset(g, 0, sizeof(g));
        for(int i = 0; i < n; i++){ //雇员i对老板s
            for(int j = 0; j < n; j++){
                int s;
                scanf("%d", &s);
                s--;
                g[s][i] += -j;
            }
        }
        for(int i = 0; i < n; i++){ //老板i对雇员s
            for(int j = 0; j < n; j++){
                int s;
                scanf("%d", &s);
                s--;
                g[i][s] += -j;
            }
        }
        nx = ny = n;
        ret = km();
        double f = -ret / 2.0 / n;
        if(ca != 1) printf("\n");
        printf("Data Set %d, Best average difference: %lf\n", ca++, f);
        memset(vis, 0, sizeof(vis));
        num = 1;
        DFS(0, 0);
    }
    return 0;
}