Stone Merge (DP)

Stone merge Time limit: 1 Sec Memory Limit: MB
Submit: Solved: 22
[Submit] [Status] [Web Board] Description

Now there are n heaps of stones, and the first heap has AI stones. Now to merge these stones into a pile, each can only merge adjacent two, the cost of each merger is the total number of stones of two piles. The minimum cost of merging all stones.

Input

The first line contains an integer T (t<=50) that represents the number of data groups.
The first row of each set of data contains an integer n (2<=n<=100) that represents the heap number of the stone.
The second line contains n positive integer ai (ai<=100), which represents the number of stones per heap.

Output

Only one row per group of data represents the minimum consolidation cost.

Sample Input

241 2 3 453 5 2 1 4

Sample Output

1933

HINT

#include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <queue > #include <map> #include <set> #include <vector> #include <math.h> #include <algorithm >using namespace std, #define LS 2*i#define rs 2*i+1#define up (i,x,y) for (i=x;i<=y;i++) #define DOWN (i,x,y) for (i=x; i>=y;i--) #define MEM (a,x) memset (A,x,sizeof (a)) #define W (a) while (a) #define LL long longconst double pi = acos (-1.0); # Define Len 200005#define mod 19999997const int INF = 0x3f3f3f3f; #define EXP 1e-6 int sum[105],dp[105][105],a[105];    int main () {int t,n,i,j,k,l;    scanf ("%d", &t);        W (t--) {scanf ("%d", &n);        MEM (sum,0);        MEM (dp,0);            Up (i,1,n) {scanf ("%d", &a[i]);        Sum[i]=sum[i-1]+a[i];                } up (L,1,n) {up (i,1,n-l) {j = i+l;                Dp[i][j]=inf;                int tem = SUM[J]-SUM[I-1]; Up (K,I,J)               Dp[i][j]=min (Dp[i][j],dp[i][k]+dp[k+1][j]+tem);    }} printf ("%d\n", Dp[1][n]);    }}/************************************************************** problem:1592 user:aking2015 language:c++ result:accepted time:108 Ms memory:1528 kb****************************************************************/

Stone Merge (DP)