HDU3359 Special Fish （KM匹配）

来源：互联网发布：免费网络拍卖平台编辑：程序博客网时间：2024/05/21 17:53

题意：告诉任意两个鱼之间的关系，然后，两条有关系的鱼的权值异或就是 spawn的值，求所有spawn和的最大值。
思路：预处理好任意两条鱼的spawn的值，KM匹配一下即可。

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<set>#include<map>#include<string>#include<cstring>#include<stack>#include<queue>#include<vector>#include<cstdlib>#define lson (rt<<1),L,M#define rson (rt<<1|1),M+1,R#define M ((L+R)>>1)#define cl(a,b) memset(a,b,sizeof(a));#define LL long long#define P pair<int,int>#define X first#define Y second#define pb push_back#define fread(zcc)  freopen(zcc,"r",stdin)#define fwrite(zcc) freopen(zcc,"w",stdout)using namespace std;const int maxn=105;const int inf=999999999;int w[maxn][maxn];int linker[maxn],lx[maxn],ly[maxn],slack[maxn];bool visx[maxn],visy[maxn];int nx,ny;bool dfs(int x){    visx[x]=true;    for(int y=0;y<ny;y++){        if(visy[y])continue;        int tmp=lx[x]+ly[y]-w[x][y];        if(tmp==0){            visy[y]=true;            if(linker[y]==-1||dfs(linker[y])){                linker[y]=x;                return true;            }        }else if(slack[y]>tmp){            slack[y]=tmp;        }    }    return false;}int km(){    cl(linker,-1);    cl(ly,0);    for(int i=0;i<nx;i++){        lx[i]=-inf;        for(int j=0;j<ny;j++)if(w[i][j]>lx[i]){            lx[i]=w[i][j];        }    }    for(int x=0;x<nx;x++){        fill(slack,slack+ny+1,inf);        while(true){            cl(visx,false);            cl(visy,false);            if(dfs(x))break;            int d=inf;            for(int i=0;i<ny;i++)if(!visy[i]&&slack[i]<d){                d=slack[i];            }            for(int i=0;i<nx;i++)if(visx[i]){                lx[i]-=d;            }            for(int i=0;i<ny;i++)if(visy[i])ly[i]+=d;            else slack[i]-=d;        }    }    int ans=0;    for(int i=0;i<ny;i++)if(linker[i]!=-1){        ans+=w[linker[i]][i];    }    return ans;}int a[maxn];char s[maxn][maxn];int main(){    int n;    while(scanf("%d",&n)&&n){        for(int i=0;i<n;i++){            scanf("%d",&a[i]);        }        for(int i=0;i<n;i++){            for(int j=0;j<n;j++){                w[i][j]=a[i]^a[j];            }        }        for(int i=0;i<n;i++){            scanf("%s",s[i]);            for(int j=0;j<n;j++){                if(s[i][j]=='0'){                    w[i][j]=0;                }            }        }        nx=ny=n;        printf("%d\n",km());    }    return 0;}

0 0