POJ-3416-Crossing-(树状数组)
来源:互联网 发布:淘宝有的不能极速退款 编辑:程序博客网 时间:2024/06/03 06:52
Wintokk has collected a huge amount of coins at THU. One day he had all his coins fallen on to the ground. Unfortunately, WangDong came by and decided to rob Wintokk of the coins.
They agreed to distribute the coins according to the following rules:
Consider the ground as a plane. Wintokk draws a horizontal line on the plane and then WangDong draws a vertical one so that the plane is divided into 4 parts, as shown below.
Wintokk will save the coins in I and III while those fit in II and IV will be taken away by the robber WangDong.
For fixed locations of the coins owned by Wintokk, they drew several pairs of lines. For each pair, Wintokk wants to know the difference between the number of the saved coins and that of the lost coins.
It's guaranteed that all the coins will lie on neither of the lines drew by that two guys.
210 329 2217 1418 233 156 2830 274 126 78 011 212 255 1019 2410 528 182 296 513 1220 2715 2611 923 2510 022 2416 3014 317 218 17 4
64422444
题意:给n个点,可以用一个有确定原点的坐标系将其划分为4个象限的点,1,3象限的点求个数和,2,4象限的点求个数和,最后求两者差的绝对值,对于每个点阵还可以有多种划分。
先将point(原点)和coin(硬币)按先x后y升序排列,这里要用两个树状数组l,r分别维护原点左右两边点,即原点左边的点都放在l,右边的点都放在r,这里将所有点先都存放入r,循环point时逐渐从r取出放入l,因为是排好序的那么处理完一个原点时,原点左边的点都在l,右边的点都在r,那么:
第一象限: sum(r,maxy)-sum(r,point[i].y)
第三象限: sum(l,point[i].y);
第二象限: sum(l,maxy)-sum(l,point[i].y)
第四象限: sum(r,point[i].y);
#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;#define M1 500005#define M2 50005int l[M1],r[M1];int maxy;struct node { int x,y,id; bool operator < (const node &obj) const { return x<obj.x||(x==obj.x&&y<obj.y); }}point[M2],coin[M2];inline int lowbit(int i){ return i&(-i);}void add(int *t,int pos,int v){ while(pos<=maxy) { t[pos]+=v; pos+=lowbit(pos); }}int sum(int *t,int i){ int res=0; while(i>0) { res+=t[i]; i-=lowbit(i); } return res;}int main(){ int T,n,m,i; int ans[M2]; scanf("%d",&T); while(T--) { memset(l,0,sizeof(l)); memset(r,0,sizeof(r)); scanf("%d%d",&n,&m); maxy=-1; for(i=1;i<=n;i++) { scanf("%d%d",&coin[i].x,&coin[i].y); coin[i].x++; coin[i].y++;//树状数组下标从1开始 if(maxy<coin[i].y) maxy=coin[i].y; } sort(coin+1,coin+1+n); for(i=1;i<=m;i++) { scanf("%d%d",&point[i].x,&point[i].y); point[i].x++; point[i].y++; point[i].id=i; if(maxy<point[i].y) maxy=point[i].y; } sort(point+1,point+1+m); for(i=1;i<=n;i++) add(r,coin[i].y,1); int it=1; for(i=1;i<=m;i++) { while(point[i].x>=coin[it].x&&it<=n) { add(l,coin[it].y,1); add(r,coin[it].y,-1); it++; } int one_three=sum(r,maxy)-sum(r,point[i].y)+sum(l,point[i].y); int two_four=sum(l,maxy)-sum(l,point[i].y)+sum(r,point[i].y); ans[point[i].id]=abs(one_three-two_four); } for(i=1;i<=m;i++) printf("%d\n",ans[i]); if(T) printf("\n"); } return 0;}
- POJ-3416-Crossing-(树状数组)
- POJ 3416 Crossing(树状数组)
- [树状数组] poj 3416 Crossing
- POJ 3416 Crossing(树状数组)
- POJ 3416——Crossing(两个树状数组,poj2352)
- POJ 3416 Crossing --离线+树状数组
- POJ-3416 Crossing (树状数组 离线处理)
- Poj 3416 Crossing + Poj 2464 Brownie Points II (树状数组)
- Gym 100463A Crossing Number(树状数组)
- POJ 3416 Crossing
- POJ 3416 Crossing
- poj 2481(树状数组)
- poj 2299(树状数组)
- poj 1195(树状数组)
- poj 2481(树状数组)
- poj 3067(树状数组)
- poj 2029(树状数组)
- poj 2155(树状数组)
- jQuery.extend()的实现方式详解及实例
- 装饰器
- 拉格朗日插值法学习小记
- 一对多 多对一 多对多
- io流使用static_cast转换输出字符串地址
- POJ-3416-Crossing-(树状数组)
- %#o和%#x的用途
- CSRF与XSS
- Android性能优化之--ViewStub
- spring/springmvc 面试题
- JavaWeb Servelt
- Windows下Zookeeper环境搭建
- JAVA中JAVA_HOME、Path、CLASSPATH的作用?
- ==与equals