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 = 16666;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define root 0, 16000, 1
int cov[N << 2];

void up(int rt) {
    if (~cov[rt << 1] && cov[rt << 1] == cov[rt << 1 | 1]) cov[rt] = cov[rt << 1];
    else cov[rt] = -1;
}

void down(int rt) { if (~cov[rt]) cov[rt << 1] = cov[rt << 1 | 1] = cov[rt];}

void build(int L, int R, int l, int r, int rt) {
    if (l >= r) {
        cov[rt] = L <= l && r <= R ? 1 : 0;
        return ;
    }
    int m = l + r >> 1;
    build(L, R, lson);
    build(L, R, rson);
    up(rt);
}

set<int> edge[N >> 1];

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 >> 1;
    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 >> 1];

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 = 1; i <= n; i++) {
            scanf("%d%d%d", &seg[i].l, &seg[i].r, &seg[i].p);
            seg[i].l <<= 1, seg[i].r <<= 1;
        }
        sort(seg + 1, seg + n + 1, cmp);
        edge[1].clear();
        build(seg[1].l, seg[1].r, root);
        for (int i = 2; i <= n; i++) {
            edge[i].clear();
            update(i, seg[i].l, seg[i].r, root);
        }
        set<int>::iterator si, sj;
        int ans = 0;
        for (int i = 1; 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
", ans);
    }
    return 0;
}

——written by Lyon