Wiki OI 1048 Stone Merge

Title Link: http://wikioi.com/problem/1048/

Algorithms and Ideas:

Although it is a relatively simple DP, the introduction of dynamic transfer equation to achieve the card, after all, is a rookie AH.

Dp[i][j] = min (Dp[i][k] + dp[k + 1][j] + s[j]-s[i-1]) k belongs to [I, j];

The process boundary of DP is important,

Maintain a sum[i] array in the input phase to represent the first and the previous I items of the stone,

DP[I][J] means to merge the minimum cost of i~j heap stones, traversing the length of the gravel, Len from 2~n,

The possible value of I is 1~n-len+1,j nature equals i+len-1,

Define k as any heap of stones in I~j, Traverse K to find the minimum cost is dp[i][j];

#include <stdio.h>
#include <string.h>
#define INF 0xffffff
int main ()
{
	int n, num[ 111], sum[111] = {0}, min;
	int dp[111][111];
	memset (DP, 0, sizeof (DP)); 
	scanf ("%d", &n);
	for (int i = 1; I <= n; i++)
	{
		scanf ("%d", &num[i]);
		Sum[i] = Sum[i-1] + num[i];
	}
	for (int len = 2, Len <= N; len++)
	{for
	    (int i = 1; I <= N-len + 1; i++)
	    {
	    	Int j = i + len-1;
	    	min = INF;
	        for (int k = i; k < J; k++)
	        {
	        	if (Dp[i][k] + dp[k + 1][j] + sum[j]-sum[i-1] < min)
	        	    min = dp[i][k] + Dp[k + 1][j] + sum[j]-sum[i-1];
	        }
	        Dp[i][j] = min;
		}
	}
	printf ("%d\n", Dp[1][n]);
	return 0;
}