hdu 3395 Special Fish

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=3395

題目大意：求交配後代 的最大值，注意每隻魚最多攻擊別的一次和被攻擊一次

題目思路：可能交配的魚之間連邊，用最大權匹配，KM算法

原題

Special Fish

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 893    Accepted Submission(s): 336

Problem Description

There is a kind of special fish in the East Lake where is closed to campus of Wuhan University. It’s hard to say which gender of those fish are, because every fish believes itself as a male, and it may attack one of some other fish who is believed to be female by it.
A fish will spawn after it has been attacked. Each fish can attack one other fish and can only be attacked once. No matter a fish is attacked or not, it can still try to attack another fish which is believed to be female by it.
There is a value we assigned to each fish and the spawns that two fish spawned also have a value which can be calculated by XOR operator through the value of its parents.
We want to know the maximum possibility of the sum of the spawns.

 

Input

The input consists of multiply test cases. The first line of each test case contains an integer n (0 < n <= 100), which is the number of the fish. The next line consists of n integers, indicating the value (0 < value <= 100) of each fish. The next n lines, each line contains n integers, represent a 01 matrix. The i-th fish believes the j-th fish is female if and only if the value in row i and column j if 1.
The last test case is followed by a zero, which means the end of the input.

 

Output

Output the value for each test in a single line.

 

Sample Input

3 1 2 3 011 101 110 0

 

Sample Output

6

 

代碼：

#include<iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn=105,OO=2147483647;
int w[maxn][maxn];
int lx[maxn],ly[maxn];
int linky[maxn];
bool visx[maxn],visy[maxn];
int slack[maxn];
int N;
bool find(int x)
{
    visx[x]=true;
    for(int y=0; y<N; ++y)
    {
        if(visy[y])continue;
        int t=lx[x]+ly[y]-w[x][y];
        if(t==0)
        {
            visy[y]=true;
            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));
    memset(lx,-1,sizeof(lx));
    memset(ly,0,sizeof(ly));
    for(int i=0; i<N; ++i)
        for(int j=0; j<N; ++j)
            if(w[i][j]>lx[i])
                lx[i]=w[i][j];
    for(int x=0; x<N; ++x)
    {
        for(int i=0; i<N; ++i)
            slack[i]=OO;
        for(;;)
        {
            memset(visx,false,sizeof(visx));
            memset(visy,false,sizeof(visy));
            if(find(x))break;
            int d=OO;
            for(int i=0; i<N; ++i)
            {
                if(!visy[i])
                    if(d>slack[i])
                        d=slack[i];
            }
            for(int i=0; i<N; ++i)
            {
                if(visx[i])
                    lx[i]-=d;
            }
            for(int i=0; i<N; ++i)
            {
                if(visy[i])
                    ly[i]+=d;
                else
                    slack[i]-=d;
            }
        }
    }
    int res=0;
    for(int j=0; j<N; ++j)
    {
        if(linky[j]!=-1)
        res+=w[linky[j]][j];
    }
    return res;
}
int main()
{
    int vl[maxn];
    char ch;
    while(scanf("%d",&N))
    {
        if(N==0)
        break;
        memset(w,0,sizeof(w));
        for(int i=0;i<N;i++)
        scanf("%d",&vl[i]);
        for(int i=0;i<N;i++)
        {
            for(int j=0;j<N;j++)
            {
                cin>>ch;
                if(ch=='1')
                {
                    w[i][j]=vl[i]^vl[j];
                }
            }
        }
    printf("%d\n",KM());
    }
}