POJ 1436 Horizontally Visible Segments (线段树)

Horizontally Visible Segments

Time Limit: 5000MSMemory Limit: 65536KTotal Submissions: 2829Accepted: 1067

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

150 4 40 3 13 4 20 2 20 2 3

Sample Output

1

线段树成段更新问题。此题跟poj2528是类似的，把坐标平面逆时针转九十度就跟那题贴海报差不多了。只不过这题问的是每三条线段是否能彼此直接用水平线相连，所以要枚举一下。做法是先按x坐标从小到大排序，每列举一条线段就判断该区间是否已经被覆盖了（既前面是否有线段和这个区间有交集，有交集既说明可以相连，这也是为什么x坐标要升序排序的原因）。然而像样例一样，0 2 2和3 4 2这两条线段中，中间的缝隙是区间（2，3），这个区间不能用整数表示了，处理方法是将每一条的纵坐标都乘以2，这样(2,3)就变成(4,6)，中间多了个整数5，这就能用线段树操作了。

邻接矩阵：60000k+ 2000ms+

邻接表：720k 125ms

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#define SIZE 8008#define ls  l,mid,rt<<1#define rs  mid+1,r,rt<<1|1using namespace std;struct node{    int l,r,x;}line[SIZE];struct Node{    int to,next;}edge[SIZE<<3];int head[SIZE],idx;int T,N,maxi;int cv[SIZE<<3];int vis[SIZE];bool cmp(node a,node b){    return a.x < b.x;}void addNode(int from,int to){    edge[idx].to = to; edge[idx].next = head[from]; head[from] = idx++;}void pushDown(int rt){    if(cv[rt] != -1)    {        cv[rt<<1] = cv[rt<<1|1] = cv[rt];        cv[rt] = -1;    }}void update(int l,int r,int rt,int L,int R,int w){    if(L <= l && r <= R)    {        cv[rt] = w;        return;    }    pushDown(rt);    int mid = (l + r) >> 1;    if(L <= mid) update(ls,L,R,w);    if(R > mid) update(rs,L,R,w);}void query(int l,int r,int rt,int L,int R,int w){    if(cv[rt] != -1)    {        if(vis[cv[rt]] != w)        {            addNode(cv[rt],w);            vis[cv[rt]] = w;        }        return;    }    if(l == r) return;    int mid = (l + r) >> 1;    if(L <= mid) query(ls,L,R,w);    if(R > mid) query(rs,L,R,w);}int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d",&N);        maxi = 0;        for(int i=1; i<=N; i++)        {            scanf("%d%d%d",&line[i].l,&line[i].r,&line[i].x);            line[i].l *= 2;            line[i].r *= 2;            maxi = max(maxi,line[i].r);        }        sort(line+1,line+1+N,cmp);        memset(cv,-1,sizeof(cv));        memset(vis,-1,sizeof(vis));        memset(head,-1,sizeof(head));        idx = 0;        for(int i=1; i<=N; i++)        {            query(0,maxi,1,line[i].l,line[i].r,i);            update(0,maxi,1,line[i].l,line[i].r,i);        }        int ans = 0;        int to;        for(int i=1; i<=N; i++)        {            for(int j=head[i]; j!=-1; j=edge[j].next)            {                to = edge[j].to;                for(int k=head[i]; k!=-1; k=edge[k].next)                {                    for(int g=head[to]; g!=-1; g=edge[g].next)                    {                        if(edge[k].to == edge[g].to)                            ans ++;                    }                }            }        }        printf("%d\n",ans);    }}