HDU 2853 Assignment 建图的巧妙性

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 32 1 33 2 41 26 22 1 32 31 2 31 2 31 2

Sample Output

2 261 2

最近在水KM算法     只要把KM理解了    至今写到的大部分都是水题

唯独这题让我写了很长时间      看了其他人的博客才明白什么叫建图的巧妙和重要性

题意    有n家公司和m个任务       所有任务和公司都已经匹配好了    要求在尽量保留原有匹配的基础上   求得最优匹配   输出改变的数量      和   改变后增加的权值

巧妙的思路来了   首先对于所有输入的权值增大K倍    这里的K必须大于n  再对原匹配的权值都加  1   这样   原匹配的优先级将大于其他匹配  

再套模板    这时出来的结果   将是匹配权值数的K倍再加上原匹配的数目       （所以这里就体现了K大于n的原因了    若是K小于n   那么大于K的部分将可能成为原匹配数目，结果将乱套………………

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;}

这个技巧必须有阿！！！！！！！