HDU 1051 Wooden Sticks (贪心)

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14126    Accepted Submission(s): 5842

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

 

#include <stdio.h>
#include <algorithm>
using namespace std;
//1051 
typedef struct node
{
    int x,y,isUsed;
}NODE;

int cmp(NODE a, NODE b)
{
    //假设x同样。就按y降序排序
    if (a.x == b.x)
    {
        return a.y>b.y;
    }
    else
    {
        return a.x>b.x;
    }
}

int main()
{
    int T,N;
    NODE c[5001];
    while (scanf("%d", &T)!=EOF)
    {
        while (T--)
        {
            scanf("%d", &N);
            for (int i=0; i<N; i++)
            {
                scanf("%d %d", &c[i].x, &c[i].y);
                c[i].isUsed = 0;
            }
            sort(c, c+N, cmp);
            
            int time = 0;
            for (int i=0; i<N; i++)
            {
                if (c[i].isUsed == 1)
                    continue;
                int x,y;
                x = c[i].x; y = c[i].y;
                //标记当前为用过了 
                c[i].isUsed = 1;
                for (int j=i+1; j<N; j++)
                {
                    if (c[j].x<=x && c[j].y<=y && c[j].isUsed == 0)
                    {
                        x = c[j].x;
                        y = c[j].y;                        
                        c[j].isUsed = 1;
                    }
                }
                time++;
            }
            printf("%d\n", time);
        }
    }
        
    return 0;
}