HDU 1051 Wooden Sticks（貪心）

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 17320    Accepted Submission(s): 7056

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
        

題意：1，第一根木棍設定時間是1min，            2，之後處理的木棍的長度為l，權重為w，如果l1<=l，w1<=w,就不用設定時間，否則就要1min的設定時間。              求最小設定時間 題解：貪心，先根據l從小到大排序，注意，當l相同時，再根據w從小到大排序。

AC代碼：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<algorithm>
typedef long long LL;
using namespace std;
struct ac{
	int l,w;
	int flag;
}node[5005];
int cmp(ac a,ac b){
	return a.l<b.l||a.l==b.l&&a.w<b.w;
}
int main()
{
     int n,i,j,count,k,t;
     while(cin>>t){
     	while(t--){
     		count=0;
     		for( i=0;i<5005;i++)
     		node[i].flag=0;
			 cin>>n;
			 for(i=0;i<n;i++){
			 	cin>>node[i].l;
			 	cin>>node[i].w;
		      }
			 sort(node,node+n,cmp);
			 for(i=0;i<n;i++){              //貪心 
			 	if(node[i].flag==1)continue;
			 	count++; 
			 	k=i;
			 	for(j=i+1;j<n;j++){
			 		if(node[j].l>=node[k].l&&node[j].w>=node[k].w&&node[j].flag==0)
			 		{
			 			node[j].flag=1;
			 			k=j;
			 		}
			 	} 
			 }
			 printf("%d\n",count);
     	}
     }
    return 0;
}