Cutting Sticks - UVa 10003 dp

最新推荐文章于 2019-07-31 23:49:30 发布

原创最新推荐文章于 2019-07-31 23:49:30 发布 · 784 阅读

0 ·

CC 4.0 BY-SA版权

dp动态规划和递推同时被 2 个专栏收录

281 篇文章

订阅专栏

UVa

218 篇文章

订阅专栏

探讨了在给定长度的木棒上进行切割以达到最小成本的方法，通过反向求解策略，优化切割顺序来降低总费用。该文详细阐述了贪心算法不可行的情况，并提供了基于BFS算法实现的最优解决方案。

Cutting Sticks

You have to cut a wood stick into pieces. The most affordable company, The Analog Cutting Machinery, Inc. (ACM), charges money according to the length of the stick being cut. Their procedure of work requires that they only make one cut at a time.

It is easy to notice that different selections in the order of cutting can led to different prices. For example, consider a stick of length 10 meters that has to be cut at 2, 4 and 7 meters from one end. There are several choices. One can be cutting first at 2, then at 4, then at 7. This leads to a price of 10 + 8 + 6 = 24 because the first stick was of 10 meters, the resulting of 8 and the last one of 6. Another choice could be cutting at 4, then at 2, then at 7. This would lead to a price of 10 + 4 + 6 = 20, which is a better price.

Your boss trusts your computer abilities to find out the minimum cost for cutting a given stick.

Input

The input will consist of several input cases. The first line of each test case will contain a positive number l that represents the length of the stick to be cut. You can assume l < 1000. The next line will contain the number n (n < 50) of cuts to be made.

The next line consists of n positive numbers c_i ( 0 < c_i < l) representing the places where the cuts have to be done, given in strictly increasing order.

An input case with l = 0 will represent the end of the input.

Output

You have to print the cost of the optimal solution of the cutting problem, that is the minimum cost of cutting the given stick. Format the output as shown below.

Sample Input

Sample Output

The minimum cutting is 200.
The minimum cutting is 22.

题意：把一根木棒在哪些位置切开，每次的花费是切的那一小段的长度，问你最小的花费。

思路：反着求解，由一小段拼成整个木棒。另外这道题不能用贪心，应该是bfs。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int num[100],num1[100],num2[100],ans,dp[100][100];
void bfs(int l,int r)
{ int i,j,k;
  if(l==r)
  { dp[l][r]=0;
    return;
  }
  for(i=l;i<r;i++)
  { if(dp[l][i]==-1)
     bfs(l,i);
    if(dp[i+1][r]==-1)
     bfs(i+1,r);
    if(dp[l][r]==-1)
     dp[l][r]=dp[l][i]+dp[i+1][r]+num2[r]-num2[l-1];
    else
     dp[l][r]=min(dp[l][r],dp[l][i]+dp[i+1][r]+num2[r]-num2[l-1]);
  }
}
int main()
{ int n,i,j,k,len;
  while(~scanf("%d",&len) && len>0)
  { scanf("%d",&n);
    memset(dp,-1,sizeof(dp));
    for(i=1;i<=n;i++)
     scanf("%d",&num[i]);
    n++;
    num[n]=len;
    for(i=1;i<=n;i++)
    { num1[i]=num[i]-num[i-1];
      num2[i]=num1[i]+num2[i-1];
    }
    bfs(1,n);
    printf("The minimum cutting is %d.\n",dp[1][n]);
  }
}