HDU1051 Wooden Sticks 贪心

本文介绍了一种通过优化木棍加工顺序来减少机器准备时间的方法。对于给定的一堆木棍,每根木棍有不同的长度和重量,目标是找到一种排列方式使得总的准备时间最短。文章提供了一个具体的算法实现,通过先选择长度和重量最小的木棍,并尽可能地连续加工不需要额外准备时间的木棍,以此来降低总的准备时间。

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Problem Description
There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows: 


(a) The setup time for the first wooden stick is 1 minute. 
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l<=l' and w<=w'. Otherwise, it will need 1 minute for setup. 


You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).
 


Input
The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1<=n<=5000, that represents the number of wooden sticks in the test case, and the second line contains n 2 positive integers l1, w1, l2, w2, ..., ln, wn, each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.
 


Output
The output should contain the minimum setup time in minutes, one per line.
 


Sample Input


4 9 5 2 2 1 3 5 1 4 

2 2 1 1 2 2 

1 3 2 2 3 1
 


Sample Output
2
1

3


排序后,首先选择l 和w都最小的,然后尽量选择不用时间的木头,之后重新选择新的最小的木头


#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define max 5010

struct node
{
    int l,w;
};

bool cmp(node x,node y)
{
    if(x.l!=y.l) return x.l<y.l;
    return x.w<y.w;
}


int main()
{
    int t,n,w,num;;
    bool mark[max];
    node a[max];
//    freopen("F://cs.txt","r",stdin);
    scanf("%d",&t);
    while(t--)
    {
        num=0;
        memset(mark,0,sizeof(mark));
        scanf("%d",&n);
        for(int i=0;i<n;++i)
        {
            scanf("%d%d",&a[i].l,&a[i].w);
        }
        sort(a,a+n,cmp);

//        for(int i=0;i<n;++i)
//            printf("%d %d\n",a[i].l,a[i].w);

        for(int i=0;i<n;++i)
        {
            if(mark[i]) continue;
            mark[i]=1;
            w=a[i].w;
            ++num;
            for(int j=i+1;j<n;++j)
            {
                if(a[j].w>=w&&!mark[j])
                {
                    w=a[j].w;
                    mark[j]=1;
                }
            }
        }
        cout<<num<<endl;
    }
    return 0;
}


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