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F - Cutting Sticks
You have to cut a wood stick into pieces. The most a?ordable 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 di?erent selections in the order of cutting can led to di?erent 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.
#include<iostream> #include<cstdio> #include<string> #include<cstring> #include<cmath> #include<algorithm> #include<cctype> #include <fstream> #include <limits> #include <vector> #include <list> #include <set> #include <map> #include <queue> #include <stack> #include <cassert> using namespace std; int m,n; int a[60][60],dp[60][60];//a: i 到 j的长度 dp i 到j的最小答案 int in(); int solve(); int out(); int main() { while(scanf("%d",&m)&&m) { in(); solve(); out(); } return 0; } int in() { memset(dp,0,sizeof(dp)); memset(a,0,sizeof(a)); scanf("%d",&n); int t=0; for(int i=0; i<n; i++) { int tt; scanf("%d",&tt); a[i][i]=tt-t; t=tt; } a[n][n]=m-t; } int solve() { for(int cha=1; cha<=n; cha++) for(int i=0; i+cha<=n; i++)//先推出底层的答案,所以要放到外面 { int j=i+cha;//求出j的位置 for(int k=i; k<j; k++)//求出i 到j的最小答案 { int t=dp[i][k]+dp[k+1][j]+a[i][k]+a[k+1][j]; if(!dp[i][j]||t<dp[i][j]) {//cout<<'@'; dp[i][j]=t; a[i][j]=a[i][k]+a[k+1][j]; } } //cout<<'!'<<i<<j<<'\t'<<dp[i][j]<<endl; } } int out() { printf("The minimum cutting is %d.\n",dp[0][n]); }
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原文地址:http://blog.csdn.net/a894383755/article/details/51350547