UVa10003 Cutting Sticks

  考虑d(i,j)表示切割点i到j这段距离的最小花费，于是d(i,j)=min(d(i,k)+d(k,j))+a[j]-a[i] ，其中j<k<i，边界条件d(i,i)=d(i,i+1)=0,最终求d(0,n+1)，复杂度o(n^3)，可采用记忆化搜索。

/*----UVa10003 Cutting Sticks
 设d(i,j)为切割木棍(i,j)的最小费用，则d(i,j)=a[j]-a[i]+min{d(i,k)+d(k,j)} i<k<=j;最终求d(0,n+1)
 为了方便，可以采用记忆化搜索
*/
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn =50+5;

int n, L,curr;
int dp[maxn][maxn],a[maxn];

int dfs(int i, int j){
	if (j - i <= 1) return  0;
	int &ans = dp[i][j];
	if (ans >= 0) return ans;
	ans = INF;
	for (int k = i + 1; k < j; k++){
		ans = min(ans, dfs(i, k) + dfs(k, j) + a[j] - a[i]);
	}
	return ans;
}
int main(){
	while (scanf("%d", &L)&&L){
		scanf("%d", &n);
		a[0] = 0, a[n + 1] = L;
		for (int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		memset(dp, -1, sizeof(dp));
		printf("The minimum cutting is %d.\n", dfs(0, n + 1));
	}
	return 0;
}