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

Problem Description

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.

 

Input

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 S_i and T_i (1 <= S_i <= T_i <= 10^9), means i-th flower will be blooming at time [S_i, T_i].
In the next M lines, each line contains an integer T_i, means the time of i-th query.

 

Output

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.

 

Sample Input

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

 

Sample Output

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:
",++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
",x,y);
        //    if(e[y-1]==q)y--;
        //    if(s[x]==q)x++;
            printf("%d
",x-y);
        }
    }
    return 0;
}