poj1436(线段树区间染色)

Horizontally Visible Segments

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

==题意:有n条垂直于x轴的线段，给出他们y范围。当两条线段之间有一条垂直于他们并且可以不经过其他线段，则可定义两条线段互相可见，求出有多少三条线段两两互相可见==

==思路:可以把每个线段看成区间，按y轴建树，然后根据线段的x位置从小到大对区间进行染色，在染色之前先对其区间内的颜色进行标记，这些颜色与当前要染的颜色一定互相可见。最后三重for循环求出两两互相可见的三条线段。但要注意有些区间如1-2，3-4，其染色之后2-3这个会被忽略，因为2-3区间没有点，而线段树储存的颜色是在点上。所以要对y扩大两倍，相当于在原先的两点间在插入一点。==

#include <iostream>#include <fstream>#include <cstdio>#include <cstring>#include <queue>#include <stack>#include <vector>#include <map>#include <cmath>#include <algorithm>#include <functional>#define inf 0x7fffffffusing namespace std;typedef long long ll;const int MAXN=1e9+10;const int MAX=16100+10;int n;bool vis[MAX>>1][MAX>>1];struct Line{    int y1,y2,x;}line[MAX>>1];struct NODE{    int l,r,color;}tree[MAX*4];int cmp(Line a,Line b){return a.x<b.x;}void init(int node,int l,int r){    tree[node].l=l;    tree[node].r=r;    tree[node].color=0;    if(l==r)    return ;    int mid=(l+r)>>1;    init(node<<1,l,mid);    init(node<<1|1,mid+1,r);    }void push_down(int node){    if(tree[node].color!=-1){        tree[node<<1].color=tree[node<<1|1].color=tree[node].color;        tree[node].color=-1;            }}void updata_tree(int node,int x,int y,int num){    int l=tree[node].l;    int r=tree[node].r;    if(x<=l&&r<=y){        tree[node].color=num;        return ;    }    push_down(node);    int mid=(l+r)>>1;    if(x<=mid)  updata_tree(node<<1,x,y,num);    if(y>mid)   updata_tree(node<<1|1,x,y,num);}void query(int node,int x,int y,int flag){    if(tree[node].color!=-1){        vis[tree[node].color][flag]=1;        return ;    }    int l=tree[node].l;    int r=tree[node].r;    int mid=(l+r)>>1;    if(l==r)        return;    push_down(node);    if(x<=mid)  query(node<<1,x,y,flag);    if(y>mid)   query(node<<1|1,x,y,flag);}int main(){    //ios::sync_with_stdio(false);    #ifdef ONLINE_JUDGE    #else    freopen("in.txt","r",stdin);    #endif    int T;    cin>>T;    while(T--){        cin>>n;        for(int i=1;i<=n;i++){              scanf("%d%d%d",&line[i].y1,&line[i].y2,&line[i].x);            line[i].y1<<=1;            line[i].y2<<=1;                 }           sort(line+1,line+n+1,cmp);        memset(vis,0,sizeof(vis));        init(1,0,MAX);        for(int i=1;i<=n;i++){            query(1,line[i].y1,line[i].y2,i);            updata_tree(1,line[i].y1,line[i].y2,i);        }               int ans=0;        for(int i=1;i<=n;i++){            for(int j=1;j<=n;j++){                if(vis[i][j]){                    for(int k=1;k<=n;k++){                        if(vis[i][k]&&vis[j][k])                            ans++;                    }                }            }        }        cout<<ans<<endl;    }    return 0;}