HDU 3506 (环形石子合并)区间dp+四边形优化

Monkey Party

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1699    Accepted Submission(s): 769

Problem Description

Far away from our world, there is a banana forest. And many lovely monkeys live there. One day, SDH(Song Da Hou), who is the king of banana forest, decides to hold a big party to celebrate Crazy Bananas Day. But the little monkeys don't know each other, so as the king, SDH must do something.
Now there are n monkeys sitting in a circle, and each monkey has a making friends time. Also, each monkey has two neighbor. SDH wants to introduce them to each other, and the rules are:
1.every time, he can only introduce one monkey and one of this monkey's neighbor.
2.if he introduce A and B, then every monkey A already knows will know every monkey B already knows, and the total time for this introducing is the sum of the making friends time of all the monkeys A and B already knows;
3.each little monkey knows himself;
In order to begin the party and eat bananas as soon as possible, SDH want to know the mininal time he needs on introducing.

 

Input

There is several test cases. In each case, the first line is n(1 ≤ n ≤ 1000), which is the number of monkeys. The next line contains n positive integers(less than 1000), means the making friends time(in order, the first one and the last one are neighbors). The input is end of file.

 

Output

For each case, you should print a line giving the mininal time SDH needs on introducing.

 

Sample Input

8 5 2 4 7 6 1 3 9

 

Sample Output

105

 

Author

PerfectCai

 

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

首先题意说的好麻烦，英语渣表示这是smg，意思就是N只猴子围成一个小圈圈大家一开始只认识自己，猴王每次只能挑出两个人介绍他们认识，花费的时间为这两个的时间总和。定义如果介绍了A和B，那么A认识的人也将认识B认识的，这所要的时间是所有的A&&B认识的人的时间和，模拟下发现这就是一个环形石子合并问题。

对于环形我们将N的环化为2*N的直线处理，最后答案就是MIN{dp[i][i+N-1]  | 1<=i<=N},N*2<=2000,传统做法会T，使用四边形优化。

对其优化内涵并不是很熟悉只知道这个定理但是看不太懂证明过程只能先记着了。

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define MAX 2005
LL dp[MAX][MAX];
LL pre[MAX]={0};
int s[MAX][MAX];
int a[MAX];
const LL inf=999999999;
int main()
{
    int N,m,i,j,k;
    while(scanf("%d",&N)!=EOF){m=N*2;
        for(i=1;i<=N;++i) scanf("%d",&a[i]),a[i+N]=a[i];
        for(i=1;i<=m;++i)
        {
            pre[i]=pre[i-1]+a[i];
            dp[i][i]=0;
            s[i][i]=i;
        }
        for(int len=2;len<=m;++len)
        {
            for(i=1,j=len;j<=m;i++,j++)
            {dp[i][j]=inf;
               for(k=s[i][j-1];k<=s[i+1][j];++k)
               {

                   LL t=dp[i][k]+dp[k+1][j]+pre[j]-pre[i-1];
                   if(dp[i][j]>t){
                    dp[i][j]=t;
                    s[i][j]=k;
                   }
               }
            }
        }
        LL ans=inf;
        for(i=1;i<=N;++i)
            if(ans>dp[i][i+N-1]) ans=dp[i][i+N-1];
        printf("%lld
",ans);
    }
    return 0;
}