《Wooden Sticks》

描述
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).

输入
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.

输出
The output should contain the minimum setup time in minutes, one per line.

这题其实就是在找递增子序列。
对于这个题：思路在于先排序，再用变量标记有没有加入。

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<string>
using namespace std;
typedef long long LL;
const int N = 5005;
struct Node
{
    int st,et,ok;//ok变量标记是否加工过了
}p[N];
bool cmp(Node a,Node b)//这里排序注意，是按开始的时间升序排，当开始的时间相同时，注意要把结束时间按升序排，不然就无法实现
{
    if(a.st == b.st) return a.et < b.et;
    else return a.st < b.st;
}
int main()
{
    int t,n,i,j,maxet,time;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;++i) scanf("%d %d",&p[i].st,&p[i].et);
        for(i=0;i<n;++i) p[i].ok = 0;
        sort(p,p+n,cmp);
        time = 0;
        for(i=0;i<n-1;++i)//注意最后一个不用当做头元素开搜，在循环结束后特判，因为最后一个元素在下面的for循环根本无法进入。
        {
            if(p[i].ok != 1)//该处没有用过就从这里开搜,这个剪枝能让时间复杂度下降很多
            {
                p[i].ok = 1;
                maxet = p[i].et;
                for(j=i+1;j<n;++j)
                {
                    if(p[j].ok != 1 && p[j].et >= maxet)
                    {
                        p[j].ok = 1;
                        maxet = p[j].et;
                    }
                }
                time++;
            }
        }
        if(p[n-1].ok != 1) time++;//对最后一个元素的特判
        printf("%d\n",time);
    }
}