【解题报告】uva10003_Cutting Sticks（切割木棍, dp）

10003 - Cutting Sticks

Time limit: 3.000 seconds

  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 

100325 50 751044 5 7 80

Sample Output 

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

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Miguel Revilla 
2000-08-21

题目大意：

首先输入一个正整数l，代表木棍长度为l。然后输入n，代表有n个需要被切的位置，接下来输入n个正整数ci，代表每个木棍上被切的位置为多少，ci为递增序列。每次切木棍，花费与所切木棍长度相等的金额，求把木棍按要求切完的最小花费。

解题思路：

木棍的总端点数量为n+2个（n个被切位置加木棍首尾），用ci代表第i个端点，木棍只能在其端点处被分割。

定义状态d(i,j)，代表切割端点i到端点j的木棍的最小花费。

状态转移方程：d(i,j) = min{ d(i,k) + d(k+j) + c[j]-c[i] | i<k<j }

根据计算需求，需要按照j-i递增的顺序递推，首先升序枚举状态首尾两端点的间隔，然后枚举起点，计算所有状态，d(1,n+2)为最终解。

#include <cstdio>#include <cstring>#define min(a,b) (a<b?a:b)int l;//木棍长度int n;//切n次int c[55];//端点位置int d[55][55];//dp状态int main(){    //freopen("in.txt","r",stdin);    while(~scanf("%d",&l),l){        scanf("%d",&n);        c[1]=0;//第一个端点为0，木棍前端        for(int i=2;i<=n+1;++i) scanf("%d",&c[i]);        c[n+2]=l;//最后一个端点，木棍尾端        memset(d,0,sizeof(d));        for(int t=2;t<n+2;t++){//枚举两个端点间间隔            for(int i=1;i+t<=n+2;++i){//枚举起点                int j=i+t;//终点                for(int k=i+1;k<j;++k){//枚举端点i和j之间的端点                    if(d[i][j]==0) d[i][j]=d[i][k]+d[k][j]+c[j]-c[i];                    else d[i][j]=min(d[i][k]+d[k][j]+c[j]-c[i],d[i][j]);                }            }        }        printf("The minimum cutting is %d.\n",d[1][n+2]);    }    return 0;}