HDU5115間隔DP

問題の意味

アイデア：

長以下〜より大きい

DP [I] [J] =分（DP [I]、[J]、DP [i]は[K - 1] + DP [K +1] [J] + [K] + B [I - 1] + B [J + 1]）。

それを参照してくださいQAQノート 

ACcode 

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

using namespace std;
const int inf = 0x3f3f3f3f;

int t,n;
int a[205],b[205];
int dp[205][205];

int main()
{
	scanf("%d",&t);
	for(int T = 1;T <= t; T++){
		scanf("%d",&n);
		for(int i = 1;i <= n; i++) scanf("%d",&a[i]);
		for(int i = 1;i <= n; i++) scanf("%d",&b[i]);
		b[n + 1] = 0,b[0] = 0;//边界
		memset(dp,0,sizeof dp);//为了后续取min 可能会调用dp[x][x - 1]
		for(int i = 1;i <= n; i++){
			for(int j = i;j <= n; j++){
				dp[i][j] = inf;
			}
		}
		for(int len = 1;len <= n; len++){
			for(int i = 1;i + len - 1 <= n;i++){
				int j = len + i - 1;
				for(int k = i;k <= j; k++){
					//why not ->for(int k = i + 1;k < j; k++) ????
					//because:when(len == 1){dp[x][x]始终等于inf没有更新;}
					dp[i][j] = min(dp[i][j],dp[i][k - 1] + dp[k + 1][j] + a[k] + b[i - 1] + b[j + 1]);
					//why(a[k] + b[i - 1] + b[j + 1])?
					//dp[i][k - 1] and dp[k + 1][j]是两个已经消灭得知cost的区间 即died wolves
				}
			}
		}
		printf("Case #%d: %d\n",T,dp[1][n]);
	}
}