UVA_10003_CuttingSticks

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 rst at 2, then at 4, then at 7. This leads to a price
of 10 + 8 + 6 = 24 because the rst 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 nd out the minimum cost for cutting a given stick.
Input
The input will consist of several input cases. The rst 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 ci (0 < ci < 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
100
3
25 50 75
10
4
4 5 7 8
0
Sample Output
The minimum cutting is 200.
The minimum cutting is 22.

简单的区间dp，

dp[i][j]代表在第i剪到j剪段中剪完对应的最小权值

因此这个问题要加上布的开头和最大长度

最后求的dp[0剪][n+1剪]才是所求

#include <iostream>#include <stdio.h>using namespace std;const int M=1005;const int IN=M*50+50;int cu[M];int dp[M][M];int main(){    int len,nc;    while(1)    {        scanf("%d",&len);        if(len==0)            break;        scanf("%d",&nc);        for(int i=1;i<=nc;i++)            scanf("%d",&cu[i]);        cu[nc+1]=len;        for(int l=2;l<=nc+1;l++)            for(int st=0;st<=nc-l+1;st++)            {                int co=IN;                int le=cu[st+l]-cu[st];                for(int mi=st+1;mi<=st+l-1;mi++)                {                    co=min(co,dp[st][mi]+dp[mi][st+l]+le);                    //cout<<st<<" "<<l<<" "<<mi<<" "<<co<<endl;                }                dp[st][st+l]=co;            }        printf("The minimum cutting is %d.\n",dp[0][nc+1]);    }    return 0;}