uva 10003 Cutting Sticks
2016-05-09 08:38
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题目:
F - Cutting SticksTime Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
You have to cut a wood stick into pieces. The most affordable company, The Analog Cutting Machinery, Inc. (ACM), charges money according to the lengthof 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.
题目大意:
给一个长度为len的木棍,分别能在ai的位置上切一刀,但费用为切之前的长度.求把所有地方都切完,费用最小的费用是多少.
题目思路:
1、可以将思路转化,求n个小木棍最小费用的合并.
2、可得到:dp[i][j]=max(dp[i][k]+dp[k+1][j]+a[j]-a[i])
题目优化:
1、从树的底端求答案,再向上推,最后推出答案
程序:
#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 =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] ); }
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