UVaOJ 10003 Cutting Sticks

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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.csdn.net/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|1const 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;}

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