Problem Description

Last year a terrible earthquake attacked Sichuan province. About 300,000 PLA soldiers attended the rescue, also ALPCs. Our mission is to solve difficulty problems to optimization the assignment of troops. The assignment is measure by efficiency, which is an integer, and the larger the better.
We have N companies of troops and M missions, M>=N. One company can get only one mission. One mission can be assigned to only one company. If company i takes mission j, we can get efficiency Eij. 
We have a assignment plan already, and now we want to change some companies’ missions to make the total efficiency larger. And also we want to change as less companies as possible.

Input

For each test case, the first line contains two numbers N and M. N lines follow. Each contains M integers, representing Eij. The next line contains N integers. The first one represents the mission number that company 1 takes, and so on.
1<=N<=M<=50, 1<Eij<=10000.
Your program should process to the end of file.

Output

For each the case print two integers X and Y. X represents the number of companies whose mission had been changed. Y represents the maximum total efficiency can be increased after changing.

Sample Input

3 3
2 1 3
3 2 4
1 26 2
2 1 3
2 3
1 2 3
1 2 3
1 2

Sample Output

2 26
1 2

————————————————————————————————————————————————————

題意： 給定一左點集有 n 個點，右點集有 m 個點的二分圖，每條邊有一個權值，輸入給出一組匹配方案，現要求滿足給定關係的中的最大匹配權值在原方案中增長了多少，並求出在原方案上最少改變了多少條邊才得到這個新的匹配

思路：題目要求在原方案上改變最少的邊得到最優匹配，實質就是優先用原匹配的邊去構建最優匹配，套用模版即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define MOD 16007
#define INF 0x3f3f3f3f
#define N 1001
#define LL long long
using namespace std;
int n,m;
int G[N][N];
int Lx[N],Ly[N];
bool visX[N],visY[N];
int linkX[N],linkY[N];
bool dfs(int x){
    visX[x]=true;
    for(int y=1;y<=m;y++){
        if(!visY[y]){
            int temp=Lx[x]+Ly[y]-G[x][y];
            if(temp==0){
                visY[y]=true;
                if(linkY[y]==-1 || dfs(linkY[y])){
                    linkX[x]=y;
                    linkY[y]=x;
                    return true;
                }
            }
        }
    }
    return false;
}
void update(){
    int minn=INF;
    for(int i=1;i<=n;i++)
        if(visX[i])
            for(int j=1;j<=m;j++)
                if(!visY[j])
                    minn=min(minn,Lx[i]+Ly[j]-G[i][j]);

    for(int i=1;i<=n;i++)
        if(visX[i])
            Lx[i]-=minn;

    for(int i=1;i<=m;i++)
        if(visY[i])
            Ly[i]+=minn;
}
int KM(){
    memset(linkX,-1,sizeof(linkX));
    memset(linkY,-1,sizeof(linkY));

    for(int i=1;i<=n;i++){
        Lx[i]=Ly[i]=0;
        for(int j=1;j<=m;j++)
            Lx[i]=max(Lx[i],G[i][j]);
    }

    for(int i=1;i<=n;i++){
        while(true){
            memset(visX,false,sizeof(visX));
            memset(visY,false,sizeof(visY));

            if(dfs(i))
                break;
            else
                update();
        }
    }

    int ans=0;
    for(int i=1;i<=m;i++)
        if(linkY[i]!=-1)
            ans+=G[linkY[i]][i];

    return ans;
}
int main(){
    while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%d",&G[i][j]);
                G[i][j]=G[i][j]*(n+1);
            }
        }

        int oldVal=0;
        for(int i=1;i<=n;i++){
            int j;
            scanf("%d",&j);
            oldVal+=(G[i][j]/(n+1));
            G[i][j]++;
        }

        int ans=KM();
        printf("%d %d\n",n-ans%(n+1),ans/(n+1)-oldVal);
    }
    return 0;
}