poj1436之线段树成段更新
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Language: Default
Horizontally Visible Segments
Description
There is a number of disjoint vertical line segments in the plane. We say that two segments are horizontally visible if they can be connected by a horizontal line segment that does not have any common points with other vertical segments. Three different vertical segments are said to form a triangle of segments if each two of them are horizontally visible. How many triangles can be found in a given set of vertical segments?
Task Write a program which for each data set: reads the description of a set of vertical segments, computes the number of triangles in this set, writes the result. Input
The first line of the input contains exactly one positive integer d equal to the number of data sets, 1 <= d <= 20. The data sets follow.
The first line of each data set contains exactly one integer n, 1 <= n <= 8 000, equal to the number of vertical line segments. Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces: yi', yi'', xi - y-coordinate of the beginning of a segment, y-coordinate of its end and its x-coordinate, respectively. The coordinates satisfy 0 <= yi' < yi'' <= 8 000, 0 <= xi <= 8 000. The segments are disjoint. Output
The output should consist of exactly d lines, one line for each data set. Line i should contain exactly one integer equal to the number of triangles in the i-th data set.
Sample Input 1 5 0 4 4 0 3 1 3 4 2 0 2 2 0 2 3 Sample Output 1 |
/*
思路:对y轴进行建树,对每条线段按横轴从小到大排序,从第1条线段开始查询并更新
查询该条线段所表示的区间内不同颜色数量(即可见线段数)并且记录可见线段,然后更新该区间颜色
最后暴力求两两可见线段数量
但是注意:0,4,1 和 0,2,2 和 3,4,2这三条线段覆盖的结果是区间0~4通过线段树查找可见线段是两条,其实是3条(2~3可见另一条)
可以把查询更新的区间*2,比如上面数据变成0,8,1 和 0,4,2 和 6,8,2则4~6之间可见一条线段
*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;
const int MAX=8000+10;
int color[MAX*2<<2];//区间可见线段编号(颜色种类)
bool mark[MAX][MAX];//记录第i条线段和第j条线段可见
struct edge{
int y1,y2,x,id;
edge(){}
edge(int Y1,int Y2,int X,int ID):y1(Y1),y2(Y2),x(X),id(ID){}
bool operator <(const edge &a)const {
return x<a.x;
}
}s[MAX];
void Upchild(int n){
if(color[n]){
color[n<<1]=color[n<<1|1]=color[n];
color[n]=0;
}
}
void Update(int L,int R,int c,int n,int left,int right){
if(L<=left && right<=R){color[n]=c;return;}
Upchild(n);
int mid=left+right>>1;
if(L<=mid)Update(L,R,c,n<<1,left,mid);
if(R>mid)Update(L,R,c,n<<1|1,mid+1,right);
}
void Query(int L,int R,int id,int n,int left,int right){
if(color[n]){mark[id][color[n]]=true;return;}
if(left == right)return;
int mid=left+right>>1;
if(L<=mid)Query(L,R,id,n<<1,left,mid);
if(R>mid)Query(L,R,id,n<<1|1,mid+1,right);
}
int main(){
int t,n,y1,y2,x;
cin>>t;
while(t--){
cin>>n;
for(int i=1;i<=n;++i){
scanf("%d%d%d",&y1,&y2,&x);
s[i]=edge(y1,y2,x,i);
}
sort(s+1,s+n+1);
memset(color,0,sizeof color);
memset(mark,false,sizeof mark);
for(int i=1;i<=n;++i){
Query(s[i].y1*2,s[i].y2*2,s[i].id,1,0,MAX*2);//先查询这条线段左边可见的线段
Update(s[i].y1*2,s[i].y2*2,s[i].id,1,0,MAX*2);//更新该区间可见线段,放在Query里更新了
}
int ans=0;
for(int i=1;i<=n;++i){//暴力统计结果
for(int j=1;j<=n;++j){
if(mark[i][j])
for(int k=1;k<=n;++k){
if(mark[i][k] && mark[j][k])++ans;
}
}
}
printf("%d\n",ans);
}
return 0;
}