UVA11383 Golden Tiger Claw

题目

UVA11383 Golden Tiger Claw

做法

\(KM\)好题啊，满足所有边\(l(x)+l(y)≥w(x,y)\)(个人理解，如不对请及时留言)，这样能满足\(\sum\limits_i^n(l(x)+l(y))\)最小值

My complete code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL maxn=1e3,inf=0x3f3f3f3f;
LL n,mi;
LL lx[maxn],ly[maxn],w[maxn][maxn],mat[maxn];
bool S[maxn],T[maxn];
bool Match(LL u){
    S[u]=true;
    for(LL i=1;i<=n;++i){
        if(!T[i]){
            LL tmp(lx[u]+ly[i]-w[u][i]);
            if(!tmp){
                T[i]=true;
                if(!mat[i] || Match(mat[i])){
                    mat[i]=u;
                    return true;
                }
            }
            else
                mi=min(mi,tmp);
        }
    }return false;
}
inline void Update(){
    for(LL i=1;i<=n;++i)
        if(S[i]) lx[i]-=mi;
    for(LL i=1;i<=n;++i)
        if(T[i]) ly[i]+=mi;
}
inline void KM(){
    for(LL i=1;i<=n;++i){
        mat[i]=lx[i]=ly[i]=0;
        for(LL j=1;j<=n;++j)
            lx[i]=max(lx[i],w[i][j]);
    }
    for(LL i=1;i<=n;++i){
        while(true){
            for(LL i=1;i<=n;++i)
                S[i]=T[i]=false;
            mi=inf;
            if(Match(i)) break;
            else Update();
        }
    }
}
int main(){
    while(scanf("%d",&n)!=EOF&&n){
        for(LL i=1;i<=n;++i)
            for(LL j=1;j<=n;++j)
                cin>>w[i][j];
        KM();
        LL ans(0);
        for(LL i=1;i<=n;++i)
            ans+=lx[i];
        for(LL i=1;i<=n;++i)
            ans+=ly[i];

        printf("%d",lx[1]);
        for(LL i=2;i<=n;++i)
            printf(" %d",lx[i]);printf("\n");
        printf("%d",ly[1]);
        for(LL i=2;i<=n;++i)
            printf(" %d ",ly[i]);printf("\n");

        printf("%d\n",ans);
    }return 0;
}

原文地址：https://www.cnblogs.com/y2823774827y/p/10349305.html