题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3395
题目大意:求交配后代 的最大值,注意每只鱼最多攻击别的一次和被攻击一次
题目思路:可能交配的鱼之间连边,用最大权匹配,KM算法
原题
Special Fish
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 893 Accepted Submission(s): 336
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.
The last test case is followed by a zero, which means the end of the input.
#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());
}
}