题解：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 ) .
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
n根木头，属性：长度，重量。每次新加工时间+1，如果后根加工的木头长度和时间<=前根，不耗时
贪心：明显的先拿最长最重的开刀
#include
#include
#include
#include
#include
using namespace std;
struct node{
int le,we;
}a[5010];
bool cmp(node x,node y)
{
if(x.le==y.le) return x.we>t;
while(t--)
{
sum=0;
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i].le>>a[i].we;
sort(a+1,a+n+1,cmp);
memset(b,true,sizeof(b));
for(int i=1;i<=n;i++)
{
if(b[i])
{
sum++;b[i]=false;
int l=a[i].le,w=a[i].we;
for(int j=i+1;j<=n;j++)
{
if(a[j].le>=l&&a[j].we>=w&&b[j])
{
b[j]=false;l=a[j].le;w=a[j].we;
}
}
}
}
cout<