POJ 1065 Wooden Sticks （贪心）

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 ) .InputThe 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.OutputThe 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

贪心基础上加个id递增的条件

#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<cmath>
#include<math.h>
using namespace std;

struct node
{
int l,w;
}a[5005];

bool vis[5005];

bool cmp(node b,node c)
{
if(b.l==c.l)
return b.w<c.w;
else
return b.l<c.l;
}

int main()
{

int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d%d",&a[i].w,&a[i].l);
sort(a,a+n,cmp);
int res=0;
memset(vis,false,sizeof(vis));
int tmpl,tmpw;
for(int i=0;i<n;i++)
{
if(!vis[i])
{
vis[i]=true;
res++;
tmpl=a[i].l;
tmpw=a[i].w;
for(int j=i+1;j<n;j++)
{
if(a[j].l>=tmpl&&a[j].w>=tmpw&&!vis[j])
{
vis[j]=true;
tmpl=a[j].l;
tmpw=a[j].w;
}
}
}
}
printf("%d\n",res);
}
return 0;
}