Flowers（二分水过。。。）

Flowers

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2579 Accepted Submission(s): 1265


[align=left]Problem Description[/align]
As
is known to all, the blooming time and duration varies between
different kinds of flowers. Now there is a garden planted full of
flowers. The gardener wants to know how many flowers will bloom in the
garden in a specific time. But there are too many flowers in the garden,
so he wants you to help him.

[align=left]Input[/align]
The first line contains a single integer t (1 <= t <= 10), the number of test cases.
For
each case, the first line contains two integer N and M, where N (1
<= N <= 10^5) is the number of flowers, and M (1 <= M <=
10^5) is the query times.
In the next N lines, each line contains two integer Si and Ti (1 <= Si <= Ti <= 10^9), means i-th flower will be blooming at time [Si, Ti].
In the next M lines, each line contains an integer Ti, means the time of i-th query.

[align=left]Output[/align]
For
each case, output the case number as shown and then print M lines. Each
line contains an integer, meaning the number of blooming flowers.
Sample outputs are available for more details.

[align=left]Sample Input[/align]

2
1 1
5 10
4
2 3
1 4
4 8
1
4
6

[align=left]Sample Output[/align]

Case #1:
0
Case #2:
1
2
1

题解：类似颜色段那题，突发奇想，二分搞了搞，upper，lower那错了半天，想了一组数据才发现问题；
代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1);
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=1e5+100;
int s[MAXN],e[MAXN];
int main(){
int t,M,N,flot=0;
scanf("%d",&t);
while(t--){
scanf("%d%d",&N,&M);
for(int i=0;i<N;i++){
scanf("%d%d",&s[i],&e[i]);
}
sort(s,s+N);sort(e,e+N);
int q;
printf("Case #%d:\n",++flot);
//    for(int i=0;i<N;i++)printf("%d ",s[i]);puts("");
//    for(int i=0;i<N;i++)printf("%d ",e[i]);puts("");
while(M--){
scanf("%d",&q);
int x=upper_bound(s,s+N,q)-s;
int y=lower_bound(e,e+N,q)-e;
//printf("%d %d\n",x,y);
//    if(e[y-1]==q)y--;
//    if(s[x]==q)x++;
printf("%d\n",x-y);
}
}
return 0;
}