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：

 1 #include
 2 #include
 3 #include
 4 #include
 5 #include
 6 using namespace std;
 7 const int INF=0x3f3f3f3f;
 8 const double PI=acos(-1);
 9 #define mem(x,y) memset(x,y,sizeof(x))
10 const int MAXN=1e5+100;
11 int s[MAXN],e[MAXN];
12 int main(){
13     int t,M,N,flot=0;
14     scanf("%d",&t);
15     while(t--){
16         scanf("%d%d",&N,&M);
17         for(int i=0;i){
18             scanf("%d%d",&s[i],&e[i]);
19         }
20         sort(s,s+N);sort(e,e+N);
21         int q;
22         printf("Case #%d:\n",++flot);
23     //    for(int i=0;i24     //    for(int i=0;i
25         while(M--){
26             scanf("%d",&q);
27             int x=upper_bound(s,s+N,q)-s;
28             int y=lower_bound(e,e+N,q)-e;
29             //printf("%d %d\n",x,y);
30         //    if(e[y-1]==q)y--;
31         //    if(s[x]==q)x++;
32             printf("%d\n",x-y);
33         }
34     }
35     return 0;
36 }

 

转载于:https://www.cnblogs.com/handsomecui/p/4965107.html