MUTC 3 A - Flowers 树状数组

Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1829    Accepted Submission(s): 901

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

 

Author

BJTU

 

Source

2012 Multi-University Training Contest 3

 

Recommend

zhoujiaqi2010

 

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离散

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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

const int maxn=211111;

int n,m;
int s[maxn],t[maxn];
int b[maxn];
int tree[maxn];
int q[maxn];
int ans[maxn];
int mn,nn;

int lowbit(int x)
{
    return x&(-x);
}

void add(int x,int val)
{
    for (int i=x;i<=mn;i+=lowbit(i))
    {
        tree[i]+=val;
    }
}

int query(int x)
{
    int ret=0;
    for (int i=x;i>0;i-=lowbit(i))
    {
        ret+=tree[i];
    }
    return ret;
}

int main()
{
    int T,cnt=0;
    scanf("%d",&T);
    while (T--)
    {
        memset(tree,0,sizeof(tree));
        scanf("%d%d",&n,&m);
        nn=1;
        for (int i=1;i<=n;i++)
        {
            scanf("%d%d",&s[i],&t[i]);
            b[nn++]=s[i];
            b[nn++]=t[i];
        }
        for (int i=1;i<=m;i++)
        {
            scanf("%d",&q[i]);
            b[nn++]=q[i];
        }
        sort(b+1,b+nn);
        mn=1;
        for (int i=2;i<nn;i++)
        {
            if (b[i]!=b[i-1]) b[++mn]=b[i];
        }
        //for (int i=1;i<=mn;i++) cerr<<b[i]<<" ";cerr<<endl;
        for (int i=1;i<=n;i++)
        {
            int l=lower_bound(b+1,b+mn+1,s[i])-b;
            int r=lower_bound(b+1,b+mn+1,t[i])-b;
            //cerr<<"l="<<l<<" r="<<r<<endl;
            add(l,1);
            add(r+1,-1);
        }
        printf("Case #%d:\n",++cnt);
        for (int i=1;i<=m;i++)
        {
            int l=lower_bound(b+1,b+mn+1,q[i])-b;
            printf("%d\n",query(l));
        }
    }
    return 0;
}