POJ-1436Horizontally Visible Segments

来源：互联网发布：node express 中间件编辑：程序博客网时间：2024/06/05 04:48

能用线段树做，但是n3也过

把坐标 * 2 来区别，端点和非端点

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct edge{    int y1,y2,x,id;}E[8008];int node[8008*8];bool ok[8008][8008];inline bool cmp(edge a,edge b){    return a.x<b.x;}void push_down(int root){    if (node[root] == 0) return;    node[root*2] = node[root];    node[root*2+1] = node[root];    node[root] = 0;}void solve(int root,int xbegin,int xend,int qbegin,int qend,int id){    if (qend < xbegin||xend<qbegin) return;    if (qbegin <= xbegin && xend <= qend)    {        if (node[root]!=0)        {            ok[id][node[root]] = true;            ok[node[root]][id] = true;            node[root] = id;            return;        }        node[root] = id;    }    else        push_down(root);    int mid = (xbegin + xend) /2;    if (xbegin == xend) return;    solve(root*2,xbegin,mid,qbegin,qend,id);    solve(root*2+1,mid+1,xend,qbegin,qend,id);}int main(){    int T;    cin >> T;    int y1,y2,x;    while (T--)    {        int N;        cin >> N;        memset(node,0,sizeof(node));        memset(ok,false,sizeof(ok));        int max_y = 0;        for (int i = 1; i<=N; i++)        {            scanf("%d%d%d",&y1,&y2,&x);            E[i].y1 = y1;            E[i].y2 = y2;            E[i].x = x;            E[i].id = i;            max_y = max(max_y,y2);        }        sort(E+1,E+N+1,cmp);        int ans = 0;        for (int i = 1;i<=N;i++)        {            solve(1,0,2 * max_y,E[i].y1 * 2,E[i].y2 * 2,E[i].id);        }        for (int i = 1; i<=N; i++)            for (int j = i+1; j<=N; j++)                if (ok[i][j])                for (int k = j+1;k<=N; k++)        {            if (ok[i][j]&&ok[i][k]&&ok[j][k]) ans++;        }        cout <<ans<<endl;    }}

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