HDU 1501 Wooden Sticks (贪心算法)

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18471    Accepted Submission(s): 7527

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

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

 

AC代码：

#include
#include
#include
#include
using namespace std;


struct node
{
    int l,w,k;
}a[10001];


bool cmp(node a,node b)
{
    if(a.l==b.l) return a.w>t;
    while(t--)
    {
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i].l>>a[i].w;
            a[i].k=1;
        }
        sort(a+1,a+n+1,cmp);
        int num=n;
        int cnt=0;
        while(num)
        {
            int teml=0,temw=0;
            for(int i=1;i<=n;i++)
            {
                 if(a[i].k==1&&a[i].w>=temw&&a[i].l>=teml)
                {
                    a[i].k=0;
                    teml=a[i].l;
                    temw=a[i].w;
                    num--;
                }
            }
            cnt++;
        }
        cout<