UVa10003 - Cutting Sticks（区间DP）

Problem Link：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=944

 

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 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.

 

•题意：给出一个长度为L的木棍，以及n个切割点 。要求切割成n+1段，每切一次花费都是原始的木棍长度 ，求最小花费。

•分析：实际上我们可以看做是n+1个物品，转化成区间dp的石子合并问题。
设计状态：dp[i][j]表示切割编号为(i~j)的木棍的花费。

•dp[i][j]=min{dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1]}

•sum[i]是前i段木棍的前缀和，实际上就是分割点 
唯一需要注意就是现在有n+1个物品了（n+1堆石子）。

AC code：

#include
using namespace std;
#define INF 0x3f3f3f3f
const int maxn=55;
int l,n,c[maxn],dp[maxn][maxn],s[maxn][maxn],sum[maxn];

int main()
{
	int i,j,k,len;
	while(cin>>l)
	{
		if(l==0) break;
		cin>>n;
		sum[0]=0;
		memset(dp,INF,sizeof(dp));
		for(i=1;i<=n;i++)
		{
			cin>>c[i];
			sum[i]=c[i];
			dp[i][i]=0;
			s[i][i]=i;
		}
		sum[n+1]=l;
		dp[n+1][n+1]=0;
		s[n+1][n+1]=n+1;
		for(len=2;len<=n+1;len++)
		{
			for(i=1;i<=n+1;i++)
			{
				j=i+len-1;
				if(j>n+1) break;
				for(k=s[i][j-1];k<=s[i+1][j];k++)
				{
					//dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1]);
					if(dp[i][j]>dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1])
					{
						dp[i][j]=dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1];
						s[i][j]=k;
					}
				}
			}
		}
		cout<<"The minimum cutting is "<