标签：using strong any fast each rom highlight tar str
On the number axis, there are N lines. The two endpoints L and R of each line are integer. Give you M queries, each query contains two intervals: [L1,R1] and [L2,R2], can you count how many lines satisfy this property: L1≤L≤R1 and L2≤R≤R2?
Input
First line will be a positive integer N (1≤N≤100000) indicating the number of lines. Following the coordinates of the N lines‘ endpoints L and R will be given (1≤L≤R≤100000). Next will be a positive integer M (1≤M≤100000) indicating the number of queries. Following the four numbers L1,R1,L2 and R2 of the M queries will be given (1≤L1≤R1≤L2≤R2≤100000).
Output
For each query output the corresponding answer.
Sample Input
3 1 3 2 4 3 5 2 1 2 3 4 1 4 5 6
Sample Output
2 1
题目链接：TOJ 4105
题意就是在给你N条在X轴上的线段，求左端点在L1~R1且右端点在L2~R2的线段条数，其实这题跟NBUT上一道题很像，问你在区间L1~R1中，值在L2~R2中有几个数，只是这题在起点计数回退时可能多退几个位子，因为线段的起点和终点坐标可能有重复的，都不能算进去，因此要用while语句来操作。离线树状数组是什么个意思呢？就是满足区间减法这样原理的话就可以这样计数：在遇到起点时减去起点到询问区间起点-1的count1，在遇到计数终点时加上询问起点~询问终点的count2，这样可以发现count1其实是count2的子集，一加一减一定会把count1抵消掉，只留下刚好符合询问区间的答案了。
代码：
#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N=100010;
struct Line
{
int l,r;
bool operator<(const Line &rhs)const
{
if(l!=rhs.l)
return l<rhs.l;
return r<rhs.r;
}
};
struct query
{
int k,r1,r2,flag,id;
query(int _k=0,int _r1=0,int _r2=0,int _flag=0,int _id=0):k(_k),r1(_r1),r2(_r2),flag(_flag),id(_id){}
bool operator<(const query &rhs)const
{
return k<rhs.k;
}
};
Line line[N];
query Q[N<<1];
int T[N],ans[N];
void init()
{
CLR(T,0);
CLR(ans,0);
}
void add(int k,int v)
{
while (k<N)
{
T[k]+=v;
k+=(k&-k);
}
}
int getsum(int k)
{
int ret=0;
while (k)
{
ret+=T[k];
k-=(k&-k);
}
return ret;
}
int main(void)
{
int n,m,i;
while (~scanf("%d",&n))
{
init();
for (i=0; i<n; ++i)
scanf("%d%d",&line[i].l,&line[i].r);
sort(line,line+n);
scanf("%d",&m);
int qcnt=0;
for (i=0; i<m; ++i)
{
int l1,r1,l2,r2;
scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
Q[qcnt++]=query(l1,l2,r2,0,i);
Q[qcnt++]=query(r1,l2,r2,1,i);
}
sort(Q,Q+qcnt);
int x=0;
for (i=0; i<qcnt; ++i)
{
while (line[x].l<=Q[i].k&&x<n)
add(line[x++].r,1);
if(Q[i].flag)
ans[Q[i].id]+=getsum(Q[i].r2)-getsum(Q[i].r1-1);
else
{
while (line[x-1].l>=Q[i].k&&x-1>=0)
{
add(line[x-1].r,-1);
--x;
}
ans[Q[i].id]-=getsum(Q[i].r2)-getsum(Q[i].r1-1);
while (line[x].l<=Q[i].k&&x<n)
{
add(line[x].r,1);
++x;
}
}
}
for (i=0; i<m; ++i)
printf("%d\n",ans[i]);
}
return 0;
}
TOJ 4105 Lines Counting（离线树状数组）
标签：using strong any fast each rom highlight tar str
原文地址：http://www.cnblogs.com/Blackops/p/6388717.html
