UVA 10003 Cutting Sticks

//  [解题方法]
//  记忆化搜索（递归，子问题的结果用备忘录存起来，避免重复求解）
//  设棍子长度n，输入的c[i]是棍子上的坐标
//  dp[x][y](即dfs(x,y))表示砍c[x]到c[y]段的最小花费
//  每次砍c[x]~c[y]段的时候枚举砍的位置i
//  状态转移：dp[x][y] = min(dp[x][i] + dp[i][y] + c[y]-c[x])(x<=i<=y)
//  注：-1表示无穷大

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define LL long long
#define M 55
#define inf 0x3fffffff

int dp[M][M], c[M];

int dfs (int x, int y)
{
    if (dp[x][y] > -1)
        return dp[x][y];
    int tp = -1, i;
    for (i = x+1; i < y; i++)
    {
        int tmp = dfs(x, i) + dfs(i, y) + c[y] - c[x];
        if (tp < 0 || tmp < tp) tp = tmp;
    }
    return (dp[x][y] = tp);
}

int main()
{
    int n, m, i;
    while (cin >> n, n)
    {
        cin >> m;
        c[0] = 0;
        for (i = 1; i <= m; i++) {
            cin >> c[i];
        }
        c[m+1] = n;
        memset (dp, -1, sizeof(dp));
        for (i = 0; i <= m; i++)
            dp[i][i+1] = 0;
        cout << "The minimum cutting is " << dfs(0, m+1) << "." << endl;
    }
    return 0;
}