HDU 2853 Assignment 建圖的巧妙性

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

最近在水KM算法     只要把KM理解了    至今寫到的大部分都是水題

唯獨這題讓我寫了很長時間      看了其他人的博客才明白什麼叫建圖的巧妙和重要性

題意    有n家公司和m個任務       所有任務和公司都已經匹配好了    要求在儘量保留原有匹配的基礎上   求得最優匹配   輸出改變的數量      和   改變後增加的權值

巧妙的思路來了   首先對於所有輸入的權值增大K倍    這裏的K必須大於n  再對原匹配的權值都加  1   這樣   原匹配的優先級將大於其他匹配  

再套模板    這時出來的結果   將是匹配權值數的K倍再加上原匹配的數目       （所以這裏就體現了K大於n的原因了    若是K小於n   那麼大於K的部分將可能成爲原匹配數目，結果將亂套………………

ac code

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cmath>

using namespace std;
const int maxn=55;
const int inf=0x3f3f3f3f;
int cnum,mnum;
int linky[maxn],lx[maxn],ly[maxn];
bool visx[maxn],visy[maxn];
int g[maxn][maxn],slack[maxn];

bool dfs(int x)
{
    visx[x]=true;
    for(int y=1;y<=mnum;y++)
    {
        if(visy[y]) continue;
        int t=lx[x]+ly[y]-g[x][y];
        if(t==0)
        {
            visy[y]=true;
            if(linky[y]==-1||dfs(linky[y]))
            {
                linky[y]=x;
                return true;
            }
        }
        else if(slack[y]>t) slack[y]=t;
    }
    return false;
}

int KM()
{
    memset(linky,-1,sizeof linky);
    memset(ly,0,sizeof ly);
    for(int i=1;i<=cnum;i++)
    {
        lx[i]=-inf;
        for(int j=1;j<=mnum;j++)
            if(g[i][j]>lx[i]) lx[i]=g[i][j];
    }
    for(int x=1;x<=cnum;x++)
    {
        for(int i=1;i<=mnum;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=1;i<=mnum;i++)
                if(!visy[i]&&d>slack[i]) d=slack[i];
            for(int i=1;i<=cnum;i++)
                if(visx[i]) lx[i]-=d;
            for(int i=1;i<=mnum;i++)
                if(visy[i]) ly[i]+=d;
                else slack[i]-=d;
        }
    }
    int result=0;
    for(int i=1;i<=mnum;i++)
    {
        if(linky[i]>-1)
            result+=g[linky[i]][i];
    }
    return result;
}

int main()
{
    int eff;
    while(scanf("%d%d",&cnum,&mnum)!=EOF)
    {
        int prenum=0,m;
        for(int i=1;i<=cnum;i++)
            for(int j=1;j<=mnum;j++)
            {
                scanf("%d",&eff);
                g[i][j]=eff*100;
            }
        for(int i=1;i<=cnum;i++)
        {
            scanf("%d",&m);
            prenum+=g[i][m];
            g[i][m]++;
        }
        int ans=KM();
        printf("%d %d\n",cnum-ans%100,ans/100-prenum/100);

    }
    return 0;
}

這個技巧必須有阿！！！！！！！