UVA

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;int dp[52][52], a[52];int main(){int l, n;while(scanf("%d", &l) != EOF){if(l == 0) return 0;scanf("%d", &n);for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);}a[0] = 0;a[n + 1] = l;memset(dp, 0, sizeof(dp));for(int d = 0; d <= n; ++d){for(int i = 1; i + d <= n; ++i){dp[i][i + d] = 1e9;for(int j = i; j <= i + d; ++j){dp[i][i + d] = min(dp[i][i + d], dp[i][j - 1] + dp[j + 1][i + d] + a[i + d + 1] - a[i - 1]);}}}printf("The minimum cutting is %d.\n", dp[1][n]);}}/*题意：长度1000的木棍，50个切割点需要切割，一次只能切一个，切的代价是切点所在木棍的长度，问切完最少需要多少代价。思路:区间dp，dp[i][j]表示[i,j]之间的切点最少需要多少代价切完，然后枚举区间内所有点作为第一个切点，然后转移即可。注意在两端加一个值辅助转移。*/