HDU 5862 Counting Intersections (离散化+扫描线+树状数组)

Posted 2020-09-11 dwtfukgv

题意：给你若干个平行于坐标轴的，长度大于0的线段，且任意两个线段没有公共点，不会重合覆盖。问有多少个交点。

析：题意很明确，可是并不好做，可以先把平行与x轴和y轴的分开，然后把平行y轴的按y坐标从小到大进行排序，然后我们可以枚举每一个平行x轴的线段，

我们可以把平行于x轴的线段当做扫描线，只不过有了一个范围，每次要高效的求出相交的线段数目，可以用一个优先队列来维护平行y轴的线段的上坐标，

如果在该平行于x轴的范围就给相应的横坐标加1，这样就很容易想到是用树状数组来维护，然后每次求出最左边和最右边，相减即可，但是由于数据范围太大，

所以我们考虑离散化横坐标。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 200000 + 10;
const int mod = 1000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
  int s, e, pos;
  Node(){ }
  Node(int ss, int ee, int p) : s(ss), e(ee), pos(p) { }
};
bool cmp1(const Node &lhs, const Node &rhs){ return lhs.s < rhs.s; }
bool cmp2(const Node &lhs, const Node &rhs){ return lhs.pos < rhs.pos; }

vector<Node> row, col;
vector<int> all;

struct node{
  int id, val;
  node(int i, int v) : id(i), val(v) { }
  bool operator < (const node &p) const{
    return val > p.val;
  }
};

int sum[maxn<<2];

int lowbit(int x){ return -x&x; }
void add(int x, int val){
  while(x <= n){
    sum[x] += val;
    x += lowbit(x);
  }
}

int getSum(int x){
  int ans = 0;
  while(x){
    ans += sum[x];
    x -= lowbit(x);
  }
  return ans;
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    row.clear();  col.clear();  all.clear();
    for(int i = 0; i < n; ++i){
      int x1, x2, y1, y2;
      scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
      if(x1 == x2){
        if(y1 > y2)  swap(y1, y2);
        col.push_back(Node(y1, y2, x1));
      }
      else{
        if(x1 > x2)  swap(x1, x2);
        row.push_back(Node(x1, x2, y1));
      }
      all.push_back(x1);
      all.push_back(x2);
    }
    sort(all.begin(), all.end());
    all.erase(unique(all.begin(), all.end()), all.end());
    sort(col.begin(), col.end(), cmp1);
    sort(row.begin(), row.end(), cmp2);
    memset(sum, 0, sizeof sum);
    n <<= 1;
    LL ans = 0;
    int cnt = 0;
    priority_queue<node> pq;
    for(int i = 0; i < row.size(); ++i){
      while(cnt < col.size() && row[i].pos >= col[cnt].s){
        int id = lower_bound(all.begin(), all.end(), col[cnt].pos) - all.begin() + 1;
        pq.push(node(id, col[cnt++].e));
        add(id, 1);
      }
      while(!pq.empty() && row[i].pos > pq.top().val){
        add(pq.top().id, -1);
        pq.pop();
      }
      int l = lower_bound(all.begin(), all.end(), row[i].s) - all.begin() + 1;
      int r = lower_bound(all.begin(), all.end(), row[i].e) - all.begin() + 1;
      ans += getSum(r) - getSum(l-1);
    }
    printf("%lld\n", ans);
  }
  return 0;
}