标签：
Wooden Sticks
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 21899 |
|
Accepted: 9350 |
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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .
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 2n 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
Source
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
struct Node{
int x;
int y;
};
int cmp(Node a,Node b)
{
if(a.x==b.x) return a.y<b.y;
return a.x<b.x;
}
int main()
{
int t;
int n;
Node node[5005];
bool flag[5005]; //标志是否已经分组
while(cin>>t)
{
while(t--)
{
memset(flag,0,sizeof(flag));
cin>>n;
int ans=0;
for(int i=0;i<n;i++)
cin>>node[i].x>>node[i].y;
sort(node,node+n,cmp);
for(int i=0;i<n;i++)
{
if(flag[i]==0)
{
int p=node[i].y;
for(int j=i;j<n;j++)
{
if(flag[j]==0&&node[j].y>=p) {flag[j]=1;p=node[j].y;}
}
ans++;
}
}
cout<<ans<<endl;
}
}
}
整体思路就是，先按照长度对结构体排序，再进行遍历分组，如果满足p<y就满足分组条件，就修改标志位，这样遍历一遍，有ans个0，就有ans组，答案就是ans；
poj 1065 贪心算法
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原文地址：http://www.cnblogs.com/hellohacker/p/5824137.html
