POJ 3253 Fence Repair

每次取最小两个合并
(n^2)可过

#include <cstdio>

const int MAXN=20011;

int read(){
	int x=0, f=1;char ch=getchar();
	while(ch<'0' || ch>'9'){if(ch=='-')f=-f;ch=getchar();}
	while(ch>='0' && ch<='9'){x=x*10+(ch-'0');ch=getchar();}
	return x*f;
}

int N;
int Num[MAXN];
long long Ans=0LL;

int Minat(int l, int r){
	int ret=r;
	for(int i=l;i<r;++i){
		if(Num[i]<Num[ret])	ret=i;
	}
	return ret;
}

int main(){
	
	N=read();
	for(int i=1;i<=N;++i)	Num[i]=read();
	
	for(int i=1, j, t;i<N;++i){
		j=Minat(i, N);
		t=Num[j];Num[j]=Num[i];Num[i]=t;
		j=Minat(i+1, N);
		t=Num[j];Num[j]=Num[i+1];Num[i+1]=t;
		Num[i+1]+=Num[i];
		Ans+=(long long)(Num[i+1]);
	}
	
	printf("%lld
", Ans);
	
	return 0;
}