poj 1436 成段更新（区间覆盖）

Horizontally Visible Segments

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 2578 Accepted: 965

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

传送门：http://blog.csdn.net/zxy_snow/article/details/6829766

http://www.cnblogs.com/wuyiqi/archive/2012/02/02/2336350.html

http://blog.csdn.net/qingniaofy/article/details/7772293

注意：如 样例 中 ，0 2 2,3 4 2这两条线段，可以看到 2 3之间是没有被覆盖的，但是在线段树中我们看不到这条线段，因为 变成 浮点数了，不能处理，那么我们可以将 坐标 x2，这样就变成 4 6，中间就多出一个点 5 了，就可以判断了。。

下面这个图是我徒手画的，很丑吧，嘻嘻。。表示了偶数点代表点，奇数代表线段，遇到有线段类的题目（用线段树做）经常要考虑乘以2，表示浮点的线段。。。poj  3225这题类似

Accepted1420K141MSC++1991B

#include<cstdio>#include<cstring>#include<vector>#include<algorithm>#define FOR(i,s,t) for(int i=(s); i<(t); i++)using namespace std;const int maxn = 20000;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1struct v_seg{    int s,t;    int x;}ss[maxn];int cover[maxn<<2];int Hash[maxn];vector<int> V[maxn];void pushdown(int rt){    if(cover[rt]!=-1){        cover[rt<<1]=cover[rt<<1|1]=cover[rt];        cover[rt]=-1;    }}void update(int L,int R,int id,int l,int r,int rt){    if(L<=l&&r<=R){        cover[rt]=id;        return ;    }    pushdown(rt);    int m=(l+r)>>1;    if(L<=m) update(L,R,id,lson);    if(R>m) update(L,R,id,rson);}void query(int L,int R,int id,int l,int r,int rt){    if(cover[rt]!=-1){        if(Hash[cover[rt]]!=id){            V[cover[rt]].push_back(id);            Hash[cover[rt]]=id;        }        return ;    }    if(l==r) return ;    pushdown(rt);    int m=(l+r)>>1;    if(L<=m) query(L,R,id,lson);    if(R>m)  query(L,R,id,rson);}int cmp(v_seg a,v_seg b){    return a.x<b.x;}int main(){    int t,i,j,k,n,T,h;    scanf("%d",&T);    while(T--){        memset(cover,-1,sizeof(cover));        memset(Hash,-1,sizeof(Hash));        scanf("%d",&n);        for(i=0;i<n;i++){            scanf("%d%d%d",&ss[i].s,&ss[i].t,&ss[i].x);            ss[i].s<<=1;ss[i].t<<=1;V[i].clear();        }        sort(ss,ss+n,cmp);        for(i=0;i<n;i++){            query(ss[i].s,ss[i].t,i,0,16000,1);            update(ss[i].s,ss[i].t,i,0,16000,1);        }        FOR(i, 0, n){sort(V[i].begin(), V[i].end());}        int ans=0;        FOR(i, 0, n){int len = V[i].size();FOR(k, 0, len){FOR(j, k+1, len) // 经测试，根据POJ的数据，最内层这个计算量不到50w{int a = V[i][k];int b = V[i][j];if( binary_search(V[a].begin(), V[a].end(), b) )ans++;}}}        printf("%d\n",ans);    }    return 0;}