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;
}