Wooden Sticks（贪心）

Wooden Sticks

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 3

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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

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

和推桌子代码差不多。

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{
	int l,w;
}f[5005];
int cmp(node a,node b){
	if(a.l==b.l)
	return a.w <b.w ;
	else 
	return a.l <b.l ;
}
int main()
{
	int t ;
	scanf("%d",&t);
	while(t--){
		int n;
		scanf("%d",&n);
	//	memset(vis,0,sizeof(vis));
			for(int i=0;i<n;i++){
				scanf("%d%d",&f[i].l ,&f[i].w );
			}
		sort(f,f+n,cmp);
		int sum=0,w;
			for(int i=0;i<n;i++){
				if(f[i].w!=0){
					sum++;
					w=f[i].w;
					for(int j=i+1;j<n;j++){
					if(f[j].w>=w){
						w=f[j].w;
						f[j].w =0; 
					}
				}
			}
		}
		printf("%d\n",sum); 
	}
	return 0;
}