D - AtCoder Express 2
Time Limit: 3 sec / Memory Limit: 1000 MB
Score: 400400 points
Problem Statement
In Takahashi Kingdom, there is a east-west railroad and NN cities along it, numbered 11, 22, 33, ..., NN from west to east. A company called AtCoder Expresspossesses MM trains, and the train ii runs from City LiLi to City RiRi (it is possible that Li=RiLi=Ri). Takahashi the king is interested in the following QQ matters:
- The number of the trains that runs strictly within the section from City pipi to City qiqi, that is, the number of trains jj such that pi≤Ljpi≤Lj and Rj≤qiRj≤qi.
Although he is genius, this is too much data to process by himself. Find the answer for each of these QQ queries to help him.
Constraints
- NN is an integer between 11 and 500500 (inclusive).
- MM is an integer between 11 and 200 000200 000 (inclusive).
- QQ is an integer between 11 and 100 000100 000 (inclusive).
- 1≤Li≤Ri≤N1≤Li≤Ri≤N (1≤i≤M)(1≤i≤M)
- 1≤pi≤qi≤N1≤pi≤qi≤N (1≤i≤Q)(1≤i≤Q)
Input
Input is given from Standard Input in the following format:
NN MM QQ L1L1 R1R1 L2L2 R2R2 :: LMLM RMRM p1p1 q1q1 p2p2 q2q2 :: pQpQ qQqQ
Output
Print QQ lines. The ii-th line should contain the number of the trains that runs strictly within the section from City pipi to City qiqi.
Sample Input 1 Copy
Copy
2 3 1 1 1 1 2 2 2 1 2
Sample Output 1 Copy
Copy
3
As all the trains runs within the section from City 11 to City 22, the answer to the only query is 33.
Sample Input 2 Copy
Copy
10 3 2 1 5 2 8 7 10 1 7 3 10
Sample Output 2 Copy
Copy
1 1
The first query is on the section from City 11 to 77. There is only one train that runs strictly within that section: Train 11. The second query is on the section from City 33 to 1010. There is only one train that runs strictly within that section: Train 33.
Sample Input 3 Copy
Copy
10 10 10 1 6 2 9 4 5 4 7 4 7 5 8 6 6 6 7 7 9 10 10 1 8 1 9 1 10 2 8 2 9 2 10 3 8 3 9 3 10 1 10
Sample Output 3 Copy
Copy
7 9 10 6 8 9 6 7 8 10
题目大意:给出一个区间1-n有m辆火车,每辆火车有自己的行驶区间,每次询问有一个l,r问l,r内有多少辆火车。
解题思路:把火车看做成二维平面上的点,每次求一下区间[l,l]-[r,r]内点的个数。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
#define LL long long
#define N 500
#define lowb(x) x&-x
int c[N][N];
map<pair<int,int>,int>mp;
void add(int x,int y)
{
for(int i=x;i<=N;i+=lowb(i))
{
for(int j=y;j<=N;j+=lowb(j))
{
c[i][j]+=1;
}
}
}
int sum(int x,int y)
{
int ans=0;
for(int i=x;i;i-=lowb(i))
{
for(int j=y;j;j-=lowb(j))
{
ans+=c[i][j];
}
}
return ans;
}
int main()
{
int n,m,q;
int l,r;
cin>>n>>m>>q;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&l,&r);
add(l,r);
}
for(int i=1;i<=q;i++)
{
scanf("%d%d",&l,&r);
printf("%d\n",sum(r,r)-sum(r,l-1)-sum(l-1,r)+sum(l-1,l-1));
}
}