hdu 3506 dp+四边形不等式优化

Monkey Party

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 953    Accepted Submission(s): 418

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

 

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刚学习四边形不等式优化，可以将一类特殊的O(n^3)复杂度的dp降为O(n^2)， 

地址：http://wenku.baidu.com/link?url=BIsI68PqTBa4fCwpNjGQd8lPKRNwaWn1EYzCRvEvpq_hq6FwnC7KFg4Wmpo_Nf_9GeDm8kqoVpEzFyRekAmCqj1vBcZfk5CPxHVVgM9eqAO

定义状态dp[i][j]为使第i只到第j只猴子互相认识的最少时间，显然有 dp[i][j] =min{ dp[i][k]+dp[k+1][j], i<=k<=j-1}+(sum[j]-sum[i-1] ) (sum[i]为前i项和）。这个基本和四边形不等式可以优化的形式一致，这里的sum[j]-sum[i-1]就是等价于w[i,j]。

显然满足区间单调性w[i',j]<=w[i,j']，和四边形不等式w[i,j]+w[i',j']<=w[i,j']+w[i',j]

因此可以优化为dp[i][j] = min{dp[i][k]+dp[k+1][j], s[i][j-1]<=k<=s[i+1][j]}+w[i,j]。其中s[i,j]是区间[i,j]取得最值的决策变量（即k），并满足单调性s[i,j-1]<=s[i,j]<=s[i+1,j]。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;


#define maxn 1001
#define inf 0x3f3f3f3f

typedef long long ll;

ll dp[2*maxn][2*maxn]; //因为1和n是neighbour，即是循环的，因此在第n项后重复添加第1到n-1项
ll sum[2*maxn];
int s[2*maxn][2*maxn];
int n;

int main()
{
    while(~scanf("%d", &n)){
        memset(dp, inf, sizeof(dp));

        sum[0]= 0;
        for(int i = 1; i <= n; i++){
            scanf("%I64d", sum+i);
            sum[i] += sum[i-1];
            s[i][i] = i;
            dp[i][i] = 0;
        }

        for(int i = n+1; i <= 2*n-1; i++){
            sum[i] = sum[i-1]+(sum[i-n]-sum[i-n-1]);
            s[i][i] = i;
            dp[i][i] = 0;
        }


        for(int len = 1; len <= n-1; len++){
            for(int i = 1; i <= n; i++){
                int k = i+len;
                int t, mini = inf;
                for(int j = s[i][k-1]; j <=s[i+1][k]; j++){
                    if(dp[i][j]+dp[j+1][k]<=mini){ t = j; mini = dp[i][j]+dp[j+1][k];}
                }
                s[i][k] = t;
                dp[i][k] = mini+sum[k]-sum[i-1];

            }
        }

        ll ans = inf;
        for(int i = 1; i <= n; i++)
            ans = min(ans, dp[i][i+n-1]);
        printf("%I64d\n",ans);
    }
    return 0;
}