POJ 2400 KM最小权匹配+输出所有配对方案

Supervisor, Supervisee
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2407   Accepted: 681

Description

Suppose some supervisors each get to hire a new person for their department. There are N people to be placed in these N departments. Each supervisor interviews all N people, and ranks them according to how much she wants each of them in her department (1 being "really want" and N being "really don't want"). In turn, each of the N candidates ranks each of the supervisors as to how much that person would like to work for that supervisor (again, 1 is "really want to work for him/her" and N is "really don't want to work for him/her"). Given the scores that each supervisor has for each candidate, and the scores each candidate has for each manager, write a computer program to determine the "best match" of candidates to supervisors. The "best match" is determined by finding the distribution that leads to the highest overall (i.e. sum of) satisfaction for all people. The closer a person is to her number one choice, the better. If everyone gets their number one choice, the average difference will be 0.

Input

The first line of the input will contain a single integer greater than 0 specifying the number of test cases. 

The next line will contain a single integer value N, 0 < N < 15, representing the number of supervisors (and the number of employees - there are N supervisors and N employees). The next N lines will be the preferences of each of the N supervisors. Each line will contain N integer entries (1 through N for employees 1 through N), each separated by a space character, that represents the preferences of that supervisor from most preferred to least preferred. More specifically, the first entry on the line will represent that supervisor's first choice, the second entry her second, and so on. The next N lines will be the preferences of the N employees, in the same format as the supervisors. 

All lines of data in the input file will end with an empty line.

Output

For each test case, write the test case number (starting with 1) followed by the best average difference written to six digits of precision to the right of the decimal point. On the next line, show which best match it was (starting with 1). On the next N lines, show each supervisor (starting with 1) followed by the employee with which she was matched (1 per line). NOTE: if there is more than one best match, matches should be listed in ascending permuted order (see sample output). 

Separate each data set with an empty line.

Sample Input

2
7
1 2 3 4 5 6 7
2 1 3 4 5 6 7
3 1 2 4 5 6 7
4 1 2 3 5 6 7
5 1 2 3 4 6 7
6 1 2 3 4 5 7
7 1 2 3 4 5 6
1 2 3 4 5 6 7
2 1 3 4 5 6 7
3 1 2 4 5 6 7
4 1 2 3 5 6 7
5 1 2 3 4 6 7
6 1 2 3 4 5 7
7 1 2 3 4 5 6

2
1 2
2 1
1 2
1 2

Sample Output

Data Set 1, Best average difference: 0.000000
Best Pairing 1
Supervisor 1 with Employee 1
Supervisor 2 with Employee 2
Supervisor 3 with Employee 3
Supervisor 4 with Employee 4
Supervisor 5 with Employee 5
Supervisor 6 with Employee 6
Supervisor 7 with Employee 7

Data Set 2, Best average difference: 0.250000
Best Pairing 1
Supervisor 1 with Employee 1
Supervisor 2 with Employee 2

Source


分析:有点坑,两个矩阵的输入顺序交换了。。。但对于题目本身并不难,首先KM求出最小权值匹配,然后dfs暴力求出匹配方案,细心一点就行。。。

代码:

//复杂度O(n^3)
//其实在求最大 最小的时候只要用一个模板就行了,把边的权值去相反数即可得到另外一个.求结果的时候再去相反数即可
//邻接矩阵特别需要注意重边的问题,切记
#pragma comment(linker,"/STACK:102400000,102400000")
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <string>
#include <math.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
typedef long long ll;   //记得必要的时候改成无符号
const int maxn=20;
const int INF=1000000000;
int n,nx,ny;
int march[maxn],lx[maxn],ly[maxn],slack[maxn];    //lx,ly为顶标,nx,ny分别为x点集y点集的个数
int visx[maxn],visy[maxn],w[maxn][maxn];

int dfs(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(march[y]==-1||dfs(march[y]))
            {
                march[y]=x;
                return 1;
            }
        }
        else if(slack[y]>t)  //不在相等子图中slack 取最小的
            slack[y]=t;
    }
    return 0;
}

int KM()
{
    int i,j;
    memset(march,-1,sizeof(march));
    memset(ly,0,sizeof(ly));
    for(i=1;i<=nx;i++)            //lx初始化为与它关联边中最大的
        for(j=1,lx[i]=-INF;j<=ny;j++)
            if(w[i][j]>lx[i])
               lx[i]=w[i][j];
    for(int x=1;x<=nx;x++)
    {
        for(i=1;i<=ny;i++)slack[i]=INF;
        while(1)
        {
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(dfs(x))     //若成功(找到了增广轨),则该点增广完成,进入下一个点的增广
                break;  //若失败(没有找到增广轨),则需要改变一些点的标号,使得图中可行边的数量增加。
                        //方法为:将所有在增广轨中(就是在增广过程中遍历到)的X方点的标号全部减去一个常数d,
                        //所有在增广轨中的Y方点的标号全部加上一个常数d
            int d=INF;
            for(i=1;i<=ny;i++)
                if(!visy[i]&&d>slack[i])
                    d=slack[i];
            for(i=1;i<=nx;i++)
                if(visx[i])
                    lx[i]-=d;
            for(i=1;i<=ny;i++)  //修改顶标后,要把所有不在交错树中的Y顶点的slack值都减去d
                if(visy[i])
                    ly[i]+=d;
                else
                    slack[i]-=d;
        }
    }
    int res=0;
    for(i=1;i<=ny;i++)
        if(march[i]>-1)
            res+=w[march[i]][i];
    return res;
}

int ans,chu[maxn],js;
bool use[maxn];
void out()
{
    printf("Best Pairing %d\n",++js);
    for(int i=1;i<=n;i++)
        printf("Supervisor %d with Employee %d\n",i,chu[i]);
}
void solve(int x,int sum)
{
    if(sum<ans)return;
    if(x==n+1){
        out();
        return;
    }
    for(int i=1;i<=n;i++){
        if(use[i]==0){
            use[i]=1;
            chu[x]=i;
            solve(x+1,sum+w[x][i]);
            use[i]=0;
        }
    }
}

int main ()
{
    int i,j,x,T;
    scanf("%d",&T);
    for(int l=1;l<=T;l++){
        scanf("%d",&n);
        nx=ny=n;
        memset(w,0,sizeof(w));
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                scanf("%d",&x);
                w[x][i]-=j-1;
            }
        }
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                scanf("%d",&x);
                w[i][x]-=j-1;
            }
        }
        ans=KM();
        memset(use,0,sizeof(use));
        printf("Data Set %d, Best average difference: %.6f\n",l,-ans*0.5/n);
        js=0;
        solve(1,0);
        printf("\n");
    }
    return 0;
}




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