UVaOJ 10003 Cutting Sticks

UVaOJ挂的我心塞。。。

题目大意：有一根长度为l的木棍，要把它从给定的n个点锯断，每锯断一次所需的费用等于木棍的长度，问锯完给定的点多需的最小的花费。

区间dp，dp[i][j]为锯i点到j点所需的最小花费，状态转移方程：dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+c[j]-c[i]);

只会写记忆化搜索，递推循环顺序傻傻搞不清楚，别人写的递推方法：http://blog.youkuaiyun.com/chuan6099/article/details/8987006

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
#include<ctime>
#define mx 7490
#define LL long long 
#define mod 1000000009
#define esp 1e-12
#define y1 y1234
#define inf 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const double PI = acos(-1.0);
using namespace std;

int l, n;
int c[55];
int dp[55][55];

int dfs(int i, int j){
	if (dp[i][j] > 0)return dp[i][j];
	if (j - i == 1)return 0;
	dp[i][j] = 0xffffff;
	for (int k = i + 1; k < j; k++){
		int ans = dfs(i, k) + dfs(k, j) + c[j] - c[i];
		if (dp[i][j]>ans)dp[i][j] = ans;
	}
	return dp[i][j];
}

int main(){
	while (scanf("%d", &l) && l){
		scanf("%d", &n);
		for (int i = 1; i <= n; i++)
			scanf("%d", &c[i]);
		c[n + 1] = l;
		memset(dp, 0, sizeof(dp));
		printf("The minimum cutting is .%d\n", dfs(0, n + 1));
	}
	return 0;
}