【HDU1051】Wooden Sticks 花样贪心算法

Wooden Sticks

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

Total Submission(s) : 7   Accepted Submission(s) : 4

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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

#include<bits/stdc++.h>//懒 万能头文件 
#include<iostream>
using namespace std;
const int maxn=5001;
struct array{
	int l,w;
}a[maxn];
//对输入的木棒按照长度排序，长度相等则按重量 
int cmp(array x,array y){
	if(x.l==y.l)
		return x.w<=y.w;
	return x.l<=y.l;
}
int main(){
	int t,n;
	bool vis[maxn];//标记已遍历过的木棒 
	cin>>t;
	while(t--){
		cin>>n;
		for(i=0;i<n;i++)
			cin>>a[i].l>>a[i].w;
		sort(a,a+n,cmp);
		memset(vis,false,sizeof(vis));
		vis[0]=true;
		int flag,i,j,st=0,cnt=0; 
		while(st<n){//好好理解while循环
			++cnt;//记录子序列的个数 
			for(i=st+1,j=st,flag=1;i<n;i++){
				if(vis[i])	continue;
				if(a[j].w<=a[i].w){
					vis[i]=true;
					j=i;
				}
				else	if(flag){
					flag=0;
					st=i;//只标记未遍历的第一根木棒 
				}
			}
			if(flag)	break;//所有木棒都遍历过了 
		}
		cout<<cnt<<endl;
	}
	return 0;
}