POJ 2400 KM算法 最小权匹配 回溯输出所有最优匹配方案
很蛋疼的一题
首先输入就很蛋疼， 据网上的神牛们纷纷说题目的矩阵给反了。然后按反着来还真给过了
KM的话 由于是 n与n的匹配，所以直接取负求KM毫无压力
但是如果两边点数不等，据说会有问题
[cpp]
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#define MAXN 505
#define MAXM 555555
#define INF 1000000000
using namespace std;
int n, m, ny, nx;
int w[MAXN][MAXN];
int lx[MAXN], ly[MAXN];
int linky[MAXN];
int visx[MAXN], visy[MAXN];
int slack[MAXN], ans;
int res[MAXN], v[MAXN];
bool find(int x)
{
visx[x] = 1;
for(int y = 1; y <= ny; y++)
{
if(visy[y]) continue;
int t = lx[x] + ly[y] - w[x][y];
if(t == 0)
{
visy[y] = 1;
if(linky[y] == -1 || find(linky[y]))
{
linky[y] = x;
return true;
}
}
else if(slack[y] > t) slack[y] = t;
}
return false;
}
int KM()
{
memset(linky, -1, sizeof(linky));
for(int i = 1; i <= nx; i++) lx[i] = -INF;
memset(ly, 0, sizeof(ly));
for(int i = 1; i <= nx; i++)
for(int j = 1; j <= ny; j++)
if(w[i][j] > lx[i]) lx[i] = w[i][j];
for(int x = 1; x <= nx; x++)
{
for(int i = 1; i <= ny; i++) slack[i] = INF;
while(true)
{
memset(visx, 0, sizeof(visx));
memset(visy, 0, sizeof(visy));
if(find(x)) break;
int d = INF;
for(int i = 1; i <= ny; i++)
if(!visy[i]) d = min(d, slack[i]);
for(int i = 1; i <= nx; i++)
if(visx[i]) lx[i] -=d;
for(int i = 1; i <= ny; i++)
if(visy[i]) ly[i] += d;
else slack[i] -= d;
}
}
ans = 0;
for(int i = 1; i <= ny; i++)
if(linky[i] > 0) ans -= w[linky[i]][i];
return ans;
}
int cnt;
void dfs(int u, int sum)
{
if(sum > ans) return;
if(u > n)
{
if(sum != ans) return;
printf("Best Pairing %d\n", ++cnt);
for(int i = 1; i <= n; i++)
printf("Supervisor %d with Employee %d\n", i, res[i]);
}
else
{
for(int i = 1; i <= n; i++)
if(!v[i])
{
res[u] = i;
v[i] = 1;
dfs(u + 1, sum - w[u][i]);
v[i] = 0;
}
}
}
int main()
{
int x, y, z, e, cas = 0, T;
scanf("%d", &T);
while(T--)
{
if(cas) printf("\n");
scanf("%d", &n);
nx = n, ny = n, m = n;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
w[i][j] = 0;
for(int i = 1; i <= n; i++)
for(int j = 0; j < n; j++)
{
scanf("%d", &x);
w[x][i] -= j;
}
for(int i = 1; i <= n; i++)
for(int j = 0; j < n; j++)
{
补充：软件开发 , C++ ,
