HDU4325-Flowers

Flowers

                                                                    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                            Total Submission(s): 3290    Accepted Submission(s): 1630

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

21 15 1042 31 44 8146

 

Sample Output

Case #1:0Case #2:121

 

Author

BJTU

 

Source

2012 Multi-University Training Contest 3

 

Recommend

zhoujiaqi2010

 

题意：给你n多花的开放时间，问一个时间有多少花开放

解题思路：先将所有时间离散化，然后就是维护树状数组，区间修改求单点

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, q;int x[300009], t[300009],a[100009],l[100009],r[100009],m;int lowbit(int k){return k&-k;}void update(int k,int val){while (k <m){x[k]+=val;k += lowbit(k);}}int getsum(int k){int sum = 0;while (k){sum += x[k];k -= lowbit(k);}return sum;}int main(){int T,cas=0;scanf("%d", &T);while (T--){int cnt = 1;memset(x, 0, sizeof x);scanf("%d%d", &n,&q);for (int i = 1; i <= n; i++){scanf("%d%d", &l[i], &r[i]);t[cnt++] = l[i];t[cnt++] = r[i];}for (int i = 1; i <= q; i++) scanf("%d", &a[i]), t[cnt++] = a[i];sort(t+1, t + cnt);m = unique(t+1, t + cnt) - t;for (int i = 1; i <= n; i++){int ll = lower_bound(t+1, t + m, l[i]) - t;int rr = lower_bound(t+1, t + m, r[i]) - t;update(ll, 1); update(rr + 1, -1);}printf("Case #%d:\n", ++cas);for (int i = 1; i <= q; i++){int aa = lower_bound(t, t + m, a[i]) - t;printf("%d\n", getsum(aa));}}return 0;}