Flowers-HDU4325（区间处理+离散化）

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

 

 

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：

现在有好几朵花，给你几段时刻，每朵花只能在指定的时间开放

求在一个时刻 有几朵花开放

 

分析：

因为他是一个区间  所以要离散化，减小数据

 

直接上代码

 

 

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<vector>
using namespace std;
#define N 1100000

struct node
{
    int x,y;
}a[N];
int f[N],p[N];

int main()
{
    int T,n,m,i,t=1;
    scanf("%d",&T);
    while(T--)
    {
        memset(p,0,sizeof(p));
        memset(a,0,sizeof(a));
        memset(f,0,sizeof(f));
        int k=0;
        scanf("%d %d",&n,&m);
        for(i=0;i<n;i++)
        {
            scanf("%d %d",&a[i].x,&a[i].y);
            p[k++]=a[i].x;
            p[k++]=a[i].x-1;
            p[k++]=a[i].y;
            p[k++]=a[i].y+1;
        }
        sort(p,p+k);
        int len=unique(p,p+k)-p;
        int Max=0;
        for(i=0;i<n;i++)
        {
            int u=lower_bound(p,p+len,a[i].x)-p;
            int v=lower_bound(p,p+len,a[i].y)-p;
            f[u]++;
            f[v+1]--;
            Max=max(Max,v+1);
        }
        for(i=1;i<=Max;i++)//防止TLE
        {
            f[i]+=f[i-1];
        }
        printf("Case #%d:\n",t++);
        while(m--)
        {
            int b;
            scanf("%d",&b);
            int u=lower_bound(p,p+len,b)-p;
            printf("%d\n",f[u]);
        }
    }
    return 0;
}