HDU 1051 Wooden Sticks（LIS+sort）

最新推荐文章于 2019-06-02 18:24:13 发布

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本文探讨了一种算法问题，即如何安排不同长度和重量的木棍加工顺序以实现最短的准备时间。通过排序和动态规划的方法找到最长的不下降子序列，从而确定所需的最少时间。

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
3
5
4 9 5 2 2 1 3 5 1 4
3
2 2 1 1 2 2
3
1 3 2 2 3 1

Sample Output
2
1
3

题解：将木棍放在机器里处理，第一根需要一分钟，剩余的如果大于等于前边放入的长度和重量，就不用费时间，否则需要一分钟，计算给所需最少的时间，其实这里的时间可以看作是求最长不下降子序列的个数
和题目最小拦截系统思路一样

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int l,w;
}woo[5010];
int sy[5010];//
int cmp(node a,node b)//从小到大排序（按长度或者重量排序均可） 
{
    if(a.l!=b.l)
    return a.l<b.l;
    else return a.w<b.w;
}
int main()
{
    int n,t,i,j,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d%d",&woo[i].l,&woo[i].w);
            sort(woo+1,woo+1+n,cmp);
        sy[1]=woo[1].w;//已经将按木棒长度排序，这里只需要记录比较木棒的重量
        k=1;//初始时间为一分钟，即现在有一个所求子序列
        for(i=2;i<=n;i++)//从第二根木棒开始比较
        {
            for(j=1;j<=k;j++)
            {
                if(woo[i].w>=sy[j])
                {
                    sy[j]=woo[i].w;
                    break;
                }
            }
            if(j>k)
            {
                k++;
                sy[k]=woo[i].w;
            }
        }
        printf("%d\n",k);
    }
    return 0;
}