Nyoj 737 Stone Merging (1) (Interval dp)

Stone merger (1)

Time Limit: 1000  ms | Memory Limit: 65535  KB

Difficulty: 3

describe

    There are N piles of stones arranged in a row, and each pile has a certain number of stones. Now take N piles of stones and make them into one pile. In the process of merging, only two adjacent piles of stones can be piled into one pile at a time. The cost of each merger is the sum of the two piles of stones, and it becomes a pile after N-1 mergers. Find the total cost minimum.

enter

There are multiple sets of test data, input to the end of the file.
The first line of each set of test data has an integer n, which means there are n piles of stones.
The next line has n (0<n<200) numbers, which represent the number of n piles of stones, separated by spaces

output

Output the minimum value of the total cost, on a separate line

sample input

3
1 2 3
7
13 7 8 16 21 4 18

Sample output

9
239

#include<stdio.h>
#include<algorithm>
using namespace std;
int dp[200][200],a[200],inf=1e9,sum[200];
intmain()
{
    int n;
    while(~scanf("%d",&n))
    {
        int i,j,k;
        for(i=1; i<=n; i++)
            scanf("%d",&a[i]),sum[i]=a[i]+sum[i-1]; //sum[i] is the sum of the previous i items
        for(i=1; i<=n; i++)
            for(j=i; j<=n; j++)
                dp[i][j]=i==j?0:inf;//求最小代价，初始化正无穷 
        for(i=1; i<n; i++)  //i表示区间间隔为i
            for(j=1; j<=n-i; j++)//j表示区间左端点
                for(k=j; k<=i+j-1; k++)//k表示[i,i+j）区间中间的点
                    dp[j][i+j]=min(dp[j][i+j],dp[j][k]+dp[k+1][i+j]+sum[i+j]-sum[j-1]); //sum[i+j]-sum[j-1]为第j项到第i+j项的总和。
        printf("%d\n",dp[1][n]);
    }
    return 0;
}