Poj1436Horizontally Visible Segments线段树

这题拆点搞。比如 [1,2] [3,4]  这两段,可以有 2.2之类的点能使一条线射穿它,但是2.2这种点整数线段上搞不出来,然后就把2*2=4 ,3*2=6;这样新的线段就可以覆盖到他穿到能与他直线相连的线段上,记录下,然后三个for 枚举答案。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include<math.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn = 8000 + 10;
bool Hash[maxn][maxn];
struct Node
{
    int a; int b; int c;
}node[maxn];
int color[maxn * 10];

int cmp(const Node &a, const Node &b)
{
    return a.c < b.c;
}
void down(int rt)
{
    if (~color[rt]){
        color[rt << 1] = color[rt << 1 | 1] = color[rt];
        color[rt] = -1;
    }
}

void build(int l, int r, int rt)
{
    color[rt] = -1;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
}

void update(int L, int R, int add, int l, int r, int rt)
{
    if (L <= l&&r <= R){
        color[rt] = add; return;
    }
    down(rt);
    int mid = (l + r) >> 1;
    if (L <= mid) update(L, R, add, lson);
    if (R > mid) update(L, R, add, rson);
}

void ask(int L, int R, int add, int l, int r, int rt)
{
    if (~color[rt]){
        Hash[add][color[rt]] = Hash[color[rt]][add] = add;
        return;
    }
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (L <= mid) ask(L, R, add, lson);
    if (R > mid) ask(L, R, add, rson);
}

int main()
{
    int Icase, n;
    cin >> Icase;
    while (Icase--){
        cin >> n;
        int ans = 0;
        int Max = 0;
        memset(Hash, 0, sizeof(Hash));
        for (int i = 0; i < n; i++){
            int a; int b; int c;
            scanf("%d%d%d", &a, &b, &c);
            node[i].a = a * 2; node[i].b = b * 2; node[i].c = c;
            if (node[i].b>Max) Max = node[i].b;
        }
        sort(node, node + n, cmp);
        build(0, Max, 1);
        for (int i = 0; i < n; i++){
            ask(node[i].a, node[i].b, i, 0, Max, 1);
            update(node[i].a, node[i].b, i, 0, Max, 1);
        }
        for (int i = 0; i < n; i++){
            for (int j = i + 1; j < n; j++){
                if (!Hash[i][j]) continue;
                for (int k = j + 1; k < n; k++){
                    if (Hash[i][k] && Hash[j][k])ans++;
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
}
 
原文地址:https://www.cnblogs.com/yigexigua/p/3928280.html