UVa 10003

这道题在时间上卡了常数因子，所以用记忆化搜索的方式无法AC。常数因子来自于递归调用所带来的时间开销

记忆化搜索的代码：

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define LEN 1000
#define MMAX 99999999
#define MIN(A,B) (A<B?A:B)

int num[LEN];
int d[LEN][LEN];
int n;

int dp(int s, int e) {
	int i;
	int tmp = MMAX;
	if(-1 != d[s][e]) return d[s][e];
	for(i = 0; i < n; i++) {
		if(num[i] > s && num[i] < e) {
			tmp = MIN(tmp, dp(s, num[i])+dp(num[i], e)+(e-s));
		}
	}
	if(MMAX == tmp) {
		d[s][e] = 0;
		return 0;
	}
	d[s][e] = tmp;
	return d[s][e];
}

int main() {
	int i, len;
	int _min = MMAX;
	while(scanf("%d", &len), len != 0) {
		scanf("%d", &n);
		memset(d, -1, sizeof(d));
		for(i = 0; i < n; i++) {
			scanf("%d", &num[i]);
		}
		printf("The minimum cutting is %d.\n", dp(0, len));
	}
	return 0;
}

 

 

非记忆化搜索的代码：

#include <stdio.h>
#include <string.h>
#define MAX 100
int table[MAX][MAX];
int main()
{
    int l; 
    while(1)
    {
        scanf("%d",&l);
        if(l==0)break;
        int n,i,j,k,a[MAX],min,t;
        scanf("%d",&n);
        for(i=1;i<=n;i++)scanf("%d",&a[i]);
        a[0]=0;a[n+1]=l;
        for(i=0;i<=n;i++)table[i][i+1]=0;
        for(i=2;i<=n+1;i++)
        {
            for(j=0;i+j<=n+1;j++)
            {
                min=100000;
                for(k=j+1;k<j+i;k++)
                    if(min>table[j][k]+table[k][j+i])
                        min=table[j][k]+table[k][j+i];
                table[j][i+j]=min+a[j+i]-a[j];
            }
        }    
        printf("The minimum cutting is %d.\n",table[0][n+1]);
    }
    return 0;
}