WoodenSticks(区间贪心类题目)

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 

这个题目是贪心类题目，算是做的第二个贪心类题目了。

思路就是先把w1或者l1其中的一个从小到大排列。我是把w从小到大排列的，但是这其中可能有w相同时l也应该从小到大排列。

排完之后刚开始我只是找了一组1分钟的，意思是其他满足上述条件的一组就是一分钟了。所以那些测试数据都也能通过，但是就是wrong answer，后来找到了原因改了代码终于对了。  好、还是请大神们指点。

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;int main(){    int a;    int x[10001][2];    scanf("%d",&a);    while(a--)    {        int b;        scanf("%d",&b);        for (int number=0;number<b;number++)        {            scanf("%d %d",&x[number][0],&x[number][1]);        }        int Mix,temp;        for (int number1=0;number1<b;number1++)        {            Mix=number1;            for (int number2=number1+1;number2<b;number2++)            {               if (x[Mix][1]>x[number2][1])                    Mix=number2;            }            if (Mix!=number1)            {                temp=x[Mix][1];                x[Mix][1]=x[number1][1];                x[number1][1]=temp;                temp=x[Mix][0];                x[Mix][0]=x[number1][0];                x[number1][0]=temp;            }        }        for (int number=0;number<b-1;number++)        {            if (x[number][1]==x[number+1][1])            {                if (x[number][0]>x[number+1][0])                {                    temp=x[number][0];                    x[number][0]=x[number+1][0];                    x[number+1][0]=temp;                }            }        }        int Count=0,Max;        for (int number1=0;number1<b;number1++)        {             if (x[number1][0]!=-1)             {                 Count++;                 Max=x[number1][0];                 for (int number2=number1+1;number2<b;number2++)                 {                     if (x[number2][0]!=-1)                     {                         if (x[number2][0]>=Max)                         {                             Max=x[number2][0];                             x[number2][0]=-1;                         }                     }                 }             }        }        printf("%d\n",Count);    }    return 0;}