UVA10003---Cutting Sticks(简单区间dp)

任意两个切割点之间无法再分割
所以如果设dp[i][j] 表示从i位置切到j位置其实没有多大意义,亲测这样会TLE
dp[i][j]表示切割第i个切割点开始的位置到第j个切割点的位置结束的木棍的最小花费
时间复杂度 O(n3)

/************************************************************************* > File Name: uva10003.cpp > Author: ALex > Mail: [email protected] > Created Time: 2015年05月21日 星期四 21时21分07秒 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int dp[55][55];
int pos[55];

int main() {
    int n, l;
    while (~scanf("%d", &l), l) {
        memset(dp, inf, sizeof(dp));
        scanf("%d", &n);
        pos[0] = 0;
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &pos[i]);
            dp[i - 1][i] = 0;
        }
        pos[n + 1] = l;
        dp[n][n + 1] = 0;
        for (int i = n; i >= 0; --i) {
            for (int j = i + 1; j <= n + 1; ++j) {
                for (int k = i; k <= j; ++k) {
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + pos[j] - pos[i]);
                }
            }
        }
        printf("The minimum cutting is %d.\n", dp[0][n + 1]);
    }
}