POJ-2400 Supervisor, Supervisee 带权值匹配+枚举所有匹配情况

题意：给定两个关系矩阵，分别表示雇主和雇员的相互好感度，好感度为1最优，N最差。如果一个人与好感度为P的人匹配的话，差值为P-1，现在要求是的总共的差值最小的匹配方法，并且输出所有的匹配方案。

解法：将两两关系矩阵转化为边上的权值，然后进行一次最大匹配，最后dfs枚举输出结果，数据中给的矩阵上下颠倒了。

代码如下：

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;const int INF = 0x3f3f3f3f;int N, sum, cnt;
int mp1[20][20], mp2[20][20];
int w[20][20];
int lx[20], ly[20], sx[20], sy[20];
int match[20], slack[20];void build() {for (int i = 1; i <= N; ++i) {for (int j = 1; j <= N; ++j) {w[i][j] = -mp1[i][j]-mp2[i][j]; // 转化为最大匹配问题 
        }}
}bool 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, 0x80, 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;}}}
}void dfs(int u, int tot, int deep) {if (tot < sum) return;if (deep == N) { // 匹配数量达到N printf("Best Pairing %d\n", ++cnt);for (int i = 1; i <= N; ++i) {printf("Supervisor %d with Employee %d\n", i, match[i]);    }return;}for (int i = 1; i <= N; ++i) {if (!sy[i]) {sy[i] = 1;match[u] = i;dfs(u+1, tot+w[u][i], deep+1);sy[i] = 0;}}
}void output(int ca) {double ret = 0;sum = cnt = 0;for (int i = 1; i <= N; ++i) {ret += w[match[i]][i];sum += w[match[i]][i];}ret = fabs(-ret / (2*N));printf("Data Set %d, Best average difference: %f\n", ca, ret);memset(sy, 0, sizeof (sy));dfs(1, 0, 0);
}int main() {int T, x, ca = 0;scanf("%d", &T);while (T--) {scanf("%d", &N);for (int i = 1; i <= N; ++i) {for (int j = 0; j < N; ++j) {scanf("%d", &x);mp1[x][i] = j; // 对第x个候选人的好感度排第j 
            }}for (int i = 1; i <= N; ++i) {for (int j = 0; j < N; ++j) {scanf("%d", &x);mp2[i][x] = j;}}build();KM();output(++ca);if (T) puts("");}return 0;    
}