NYOJ 522 裸的树状数组
又是一道水题，，话说这次月赛水题真的很多很多，貌似比赛时写出来的题都是水题。。。看来，水平也就能水一下题而已。。。。不过这道题比赛时还是坑了不少人，很多人在处理边界问题0的时候没有注意，都TLE了，，当时我也TLE了一次，后来仔细想了想，改过后就ac了。相比那些一直TLE到最后的孩纸来说，我算是幸运了。不过，这道题难度有点高了，除了边界问题外，就是道裸的树状数组，没什么难度的。题目：
Interval
时间限制：
2000 ms | 内存限制：
65535 KB
难度：
4
- 描述
-
There are n(1 <= n <= 100000) intervals [ai, bi] and m(1 <= m <= 100000) queries, -100000 <= ai <= bi <= 100000 are integers.Each query contains an integer xi(-100000 <= x <= 100000). For each query, you should answer how many intervals convers xi.
- 输入
-
The first line of input is the number of test case.
For each test case,
two integers n m on the first line,
then n lines, each line contains two integers ai, bi;
then m lines, each line contains an integer xi. - 输出
- m lines, each line an integer, the number of intervals that covers xi.
- 样例输入
-
2 3 4 1 3 1 2 2 3 0 1 2 3 1 3 0 0 -1 0 1
- 样例输出
-
0 2 3 2 0 1 0
#include <iostream>
#include <cstdio>
#include <string.h>
#include <cmath>
const int N=200002;
const int M=100001;
int num[N],ans[N];
int lowbit(int x){
return x&(-x);
}
void update(int m,int value){
while(m>0){
num[m]+=value;
m-=lowbit(m);
}
}
int sum(int x){
int s=0;
while(x<=N){
s+=num[x];
x+=lowbit(x);
}
return s;
}
int main(){
//freopen("1.txt","r",stdin);
int numcase;
scanf("%d",&numcase);
while(numcase--){
memset(num,0,sizeof(num));
// memset(ans,0,sizeof(ans));
int n,m;
scanf("%d%d",&n,&m);
int x,y;
while(n--){
scanf("%d%d",&x,&y);
x=x+M;
y=y+M;
update(x-1,-1);
update(y,1);
}
while(m--){
scanf("%d",&x);
x=x+M;
//if(!ans[x])
printf("%d\n",sum(x));
/*else
printf("%d\n",ans[x]);*/
}
}
return 0;
}