poj 1436 && zoj 1391 Horizontally Visible Segments (Segment Tree)

ZOJ :: Problems :: Show Problem

1436 -- Horizontally Visible Segments

　　用线段树记录表面能被看见的线段的编号，然后覆盖的时候同时把能看到的线段记录下来。这里要用到拆点，在两个整点之间插入一个点。

　　最后O(n^2)统计三角形的个数，因为每条线段可以看见的另外的线段的条数不多，所以可以直接枚举两条边。

代码如下：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <set>

 using namespace std;

 const int N = ;
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define root 0, 16000, 1
 int cov[N << ];

 void up(int rt) {
     if (~cov[rt << ] && cov[rt << ] == cov[rt <<  | ]) cov[rt] = cov[rt << ];
     else cov[rt] = -;
 }

 void down(int rt) { if (~cov[rt]) cov[rt << ] = cov[rt <<  | ] = cov[rt];}

 void build(int L, int R, int l, int r, int rt) {
     if (l >= r) {
         cov[rt] = L <= l && r <= R ?  : ;
         return ;
     }
     int m = l + r >> ;
     build(L, R, lson);
     build(L, R, rson);
     up(rt);
 }

 set<int> edge[N >> ];

 void update(int k, int L, int R, int l, int r, int rt) {
     if (L <= l && r <= R && ~cov[rt]) {
         if (cov[rt]) edge[k].insert(cov[rt]);
         cov[rt] = k;
         return ;
     }
     int m = l + r >> ;
     down(rt);
     if (L <= m) update(k, L, R, lson);
     if (m < R) update(k, L, R, rson);
     up(rt);
 }

 struct Node {
     int l, r, p;
 } seg[N >> ];

 bool cmp(Node a, Node b) { return a.p < b.p;}

 int main() {
 //    freopen("in", "r", stdin);
     int T, n;
     scanf("%d", &T);
     while (T-- && ~scanf("%d", &n)) {
         for (int i = ; i <= n; i++) {
             scanf("%d%d%d", &seg[i].l, &seg[i].r, &seg[i].p);
             seg[i].l <<= , seg[i].r <<= ;
         }
         sort(seg + , seg + n + , cmp);
         edge[].clear();
         build(seg[].l, seg[].r, root);
         for (int i = ; i <= n; i++) {
             edge[i].clear();
             update(i, seg[i].l, seg[i].r, root);
         }
         set<int>::iterator si, sj;
         int ans = ;
         for (int i = ; i <= n; i++) {
 //            cout << i << " : ";
 //            for (si = edge[i].begin(); si != edge[i].end(); si++) cout << *si << ' ';
 //            cout << endl;
             for (si = edge[i].begin(); si != edge[i].end(); si++) {
                 int t = *si;
                 for (sj = edge[i].begin(); sj != edge[i].end(); sj++) {
                     ans += edge[t].find(*sj) != edge[t].end();
                 }
             }
         }
         printf("%d\n", ans);
     }
     return ;
 }

——written by Lyon