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