【HDOJ】【3506】Monkey Party

DP/四边形不等式

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　　裸题环形石子合并……

　　拆环为链即可

//HDOJ 3506
#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define rep(i,n) for(int i=0;i<n;++i)
#define F(i,j,n) for(int i=j;i<=n;++i)
#define D(i,j,n) for(int i=j;i>=n;--i)
#define pb push_back
#define CC(a,b) memset(a,b,sizeof(a))
using namespace std;
int getint(){
    int v=0,sign=1; char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') sign=-1; ch=getchar();}
    while(isdigit(ch))  {v=v*10+ch-'0'; ch=getchar();}
    return v*sign;
}
const int N=2010;
const int INF=1000000000;
const double eps=1e-8;
/*******************template********************/
int s[N][N],a[N];
int dp[N][N];
int main(){
#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif
    int n;
    while(scanf("%d",&n)!=EOF && n){
        CC(a,0);CC(s,0);
        F(i,1,n) a[i+n]=a[i]=getint();
        F(i,1,n<<1) a[i]+=a[i-1];
        F(i,1,n*2-1){
            dp[i][i+1]=a[i+1]-a[i-1];
            s[i][i+1]=i;
        }
        F(len,3,n)
            F(i,1,n*2-len+1){
                int j=i+len-1;
                dp[i][j]=INF;
                F(k,s[i][j-1],s[i+1][j]){
                    int tmp=dp[i][k]+dp[k+1][j]+a[j]-a[i-1];
                    if (tmp<dp[i][j]){
                        dp[i][j]=tmp;
                        s[i][j]=k;
                    }
                }
            }
        int ans=INF;
        F(i,1,n) ans=min(ans,dp[i][i+n-1]);
        printf("%d
",ans);
    }        
    return 0;
}