http://acm.hdu.edu.cn/showproblem.php?pid=4325

Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2633    Accepted Submission(s): 1290

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

题目大意：n种花，每种花的花期为[s,t]，m次查询，每次查询问ti时刻有多少种花开

数据范围过大，无法直接开数组，那么就要用到离散化处理

方法类似hdu 5124

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>

using namespace std;

typedef long long ll;
const int N = 2000010;
const int INF = 0x3f3f3f3f;

int c[2 * N], a[N], b[N], used[N], d[N];

int cmp(const void *a, const void *b)
{
    return *(int *)a - *(int *)b;
}

int main()
{
    int t, n, m, f = 0;
    int p, q, k, e;
    scanf("%d", &t);
    while(t--)
    {
        f++;
        k = 0;
        memset(used, 0, sizeof(used));
        scanf("%d%d", &n, &m);
        for(int i = 0 ; i < n ; i++)
        {
            scanf("%d%d", &p, &q);
            a[i] = p;
            b[i] = q;
            c[k++] = p;
            c[k++] = p - 1;/***///这里为嘛，个人认为是防止查询的数比q要小
            c[k++] = q;
            c[k++] = q + 1;/***/
        }

        qsort(c, k, sizeof(c[0]), cmp);

        k = unique (c, c + k)- c;//去重
        int x, y, maxn = -INF;
        for(int i = 0 ; i < n ; i++)
        {
            x = lower_bound(c, c + k, a[i]) - c;
            y = lower_bound(c, c + k, b[i]) - c;
            used[x]++;
            used[y + 1]--;
            maxn = max(maxn, y + 1);
        }

        for(int i = 0 ; i < maxn ; i++)
            used[i + 1] += used[i];

        printf("Case #%d:\n", f);
        while(m--)
        {
            scanf("%d", &e);
            int s = lower_bound(c, c + k, e) - c;

            printf("%d\n", used[s]);
        }
    }
    return 0;
}