GBus count



Problem

There exists a straight line along which cities are built.

Each city is given a number starting from 1. So if there are 10 cities, city 1 has a number 1, city 2 has a number 2,... city 10 has a number 10.

Different buses (named GBus) operate within different cities, covering all the cities along the way. The cities covered by a GBus are represented as 'first_city_number last_city_number' So, if a GBus covers cities 1 to 10 inclusive, the cities covered by it are represented as '1 10'

We are given the cities covered by all the GBuses. We need to find out how many GBuses go through a particular city.

Input

The first line contains the number of test cases (T), after which T cases follow each separated from the next with a blank line.
For each test case,
The first line contains the number of GBuses.(N)
Second line contains the cities covered by them in the form
a1 b1 a2 b2 a3 b3...an bn
where GBus1 covers cities numbered from a1 to b1, GBus2 covers cities numbered from a2 to b2, GBus3 covers cities numbered from a3 to b3, upto N GBuses.
Next line contains the number of cities for which GBus count needs to be determined (P).
The below P lines contain different city numbers.
Output

For each test case, output one line containing "Case #Ti:" followed by P numbers corresponding to the number of cities each of those P GBuses goes through.

Limits

1 <= T <= 10
ai and bi will always be integers.

Small dataset

1 <= N <= 50
1 <= ai <= 500, 1 <= bi <= 500
1 <= P <= 50

Large dataset

1 <= N <= 500
1 <= ai <= 5000, 1 <= bi <= 5000
1 <= P <= 500

Sample

Input
  
2
4
15 25 30 35 45 50 10 20
2
15
25

10
10 15 5 12 40 55 1 10 25 35 45 50 20 28 27 35 15 40 4 5
3
5
10
27

Output
 
Case #1: 2 1
Case #2: 3 3 4

解法一：

<pre class="cpp" name="code">#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<fstream>using namespace std;int count[5005];int bus[1005];int query[505];void init(){    memset(count, 0, sizeof(count));    memset(bus, 0, sizeof(bus));    memset(query, 0, sizeof(query));}int main(){    int T, N, P, index=1;    ifstream fcin;    ofstream fcout;    fcin.open("B-large-practice.in");    fcout.open("result_large.txt");    fcin>>T;    while(index<=T)    {        init();        fcin>>N;        for(int i=0;i<2*N;i++)            fcin>>bus[i];        fcin>>P;        for(int i=0;i<P;i++)            fcin>>query[i];        fcout<<"Case #"<<index<<":";        for(int i=0;i<P;i++)        {            int re=0;            for(int j=0;j<2*N;j+=2)            {                if(query[i]>=bus[j]&&query[i]<=bus[j+1])                    re++;            }            fcout<<" "<<re;        }        fcout<<endl;        index++;    }}

解法二：

线段树，区间更新。城市数量[1,5000]，在给出每条公交线路时，将[ai,bi]区间内计数+1。

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<fstream>#define LL(x) (x*2+1)#define RR(x) (x*2+2)using namespace std;const int maxb=5000;int q[505];struct node{    int left, right, num;    int mid()    {        return (int)(left+right)/2;    }}bus[4*maxb+5];void build(int left, int right, int index){    bus[index].left=left;    bus[index].right=right;    bus[index].num=0;    if(left<right)    {        int mid=bus[index].mid();        build(left, mid, LL(index));        build(mid+1, right, RR(index));    }}void insert(int l, int r, int index){    if((l==bus[index].left)&&(r==bus[index].right))    {        bus[index].num++;        return;    }    if(bus[index].left==bus[index].right)        return;    int mid=bus[index].mid();    if(r<=mid)        insert(l, r, LL(index));    else if(mid<l)        insert(l, r, RR(index));    else    {        insert(l, mid, LL(index));        insert(mid+1, r, RR(index));    }}int query(int x, int index){    if((bus[index].left==bus[index].right))    {        return bus[index].num;    }    int mid=bus[index].mid();    if(x<=mid)    {        return bus[index].num+query(x, LL(index));    }    if(x>mid)    {        return bus[index].num+query(x, RR(index));    }}int main(){    int T, N, P, index=1;    ifstream fcin;    ofstream fcout;    fcin.open("B-small-practice.in");    fcout.open("B_result_small_2.txt");    fcin>>T;    while(index<=T)    {        build(1, maxb, 0);        fcin>>N;        int x, y;        for(int i=0;i<N;i++)        {            fcin>>x>>y;            insert(x, y, 0);        }        fcin>>P;        for(int i=0;i<P;i++)            fcin>>q[i];        fcout<<"Case #"<<index<<":";        for(int i=0;i<P;i++)        {            fcout<<" "<<query(q[i], 0);        }        fcout<<endl;        index++;    }}