hdu 5596 GTW likes gt

Problem Description

Long long ago, there were n adorkable GT. Divided into two groups, they were playing games together, forming a column. The i−th GT would randomly get a value of ability bi. At the i−th second, the i−th GT would annihilate GTs who are in front of him, whose group differs from his, and whose value of ability is less than his.

In order to make the game more interesting, GTW, the leader of those GTs, would emit energy for m times, of which the i−th time of emitting energy is ci. After the ci second, b1,b2,...,bci would all be added 1.

GTW wanted to know how many GTs would survive after the n−th second.

Input

The first line of the input file contains an integer T(≤5), which indicates the number of test cases.

For each test case, there are n+m+1 lines in the input file.

The first line of each test case contains 2 integers n and m, which indicate the number of GTs and the number of emitting energy, respectively.(1≤n,m≤50000)

In the following n lines, the i−th line contains two integers ai and bi, which indicate the group of the i−th GT and his value of ability, respectively. (0≤ai≤1,1≤bi≤106)

In the following m lines, the i−th line contains an integer ci, which indicates the time of emitting energy for i−th time.

Output

There should be exactly T lines in the output file.

The i−th line should contain exactly an integer, which indicates the number of GTs who survive.
 

Sample Input

1
4 3
0 3
1 2
0 3
1 1
1
3
4

 

Sample Output

3

Hint

 

Source

BestCoder Round #66 (div.2)

 

问这个数与之后比它大的数的影响的。可以从后往前面扫，记最大值，如果碰到比最大值小的，去掉即可。

 

#include<stdio.h>  
#include<algorithm>  
#include<math.h>  
#include<queue>  
#include<string.h>  
#include<stack>  
using namespace std;  
const int  maxn = 1e5+10;  
int c[maxn];  
struct node  
{  
    int data;  
    int id;  
}a[maxn];  
int main()  
{  
    int T;  
    scanf("%d",&T);  
    while(T--)  
    {  
      int  n,m;  
      scanf("%d%d",&n,&m);  
      for(int i = 1;i<=n;i++)  
      {  
          scanf("%d%d",&a[i].id,&a[i].data);  
      }  
      int x;  
      memset(c,0,sizeof(c));  
        for(int i = 1;i<=m;i++)  
          {  
              scanf("%d",&x);  
              c[x]++;  
          }  
          int ans= n;  
          int max1= -1;  
          int max2= -1;  
          for(int i= n;i>=1;i--)  
          {    max1-=c[i];  
               max2-=c[i];  
              if(a[i].id ==1)  
               {  //max1-=c[i];  
                   if(a[i].data<max2)  
                   {  
                       ans--;  
                   }  
                   max1= max(max1,a[i].data);  
              }  
              else  
              {  
                 if(a[i].data<max1)  
                   {  
                       ans--;  
                   }  
                   max2= max(max2,a[i].data);  
              }  
          }  
          printf("%d
",ans);  
    }  
    return 0;  
}