BZOJ 1818: [Cqoi2010]内部白点 (BIT + 扫描线)

就是求多条线段的交点数,直接BIT+扫描线就行了. 注意不要算重最初存在的点.

CODE

#include<bits/stdc++.h>
using namespace std;
char cb[1<<15],*cs=cb,*ct=cb;
#define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
template<class T>inline void read(T &res) {
	char ch; int flg = 1; for(;!isdigit(ch=getc());)if(ch=='-')flg=-flg;
	for(res=ch-'0';isdigit(ch=getc());res=res*10+ch-'0'); res*=flg;
}
const int MAXN = 100005;
struct node { int x, y; }a[MAXN];
struct Node { int x, y, v; inline bool operator <(const Node &o)const { return y < o.y; } }q[MAXN];
inline bool cmpx(const node &a, const node &b) { return a.x == b.x ? a.y < b.y : a.x < b.x; }
inline bool cmpy(const node &a, const node &b) { return a.y == b.y ? a.x < b.x : a.y < b.y; }
int n, T[MAXN], b[MAXN], tot, N;
inline void upd(int x, int val) { while(x <= N) T[x] += val, x += x&-x; }
inline int qsum(int x) { int re = 0; while(x) re += T[x], x -= x&-x; return re; }
int main() {
	read(n);
	for(int i = 1; i <= n; ++i)
		read(a[i].x), read(a[i].y), b[++N] = a[i].x;
	sort(b + 1, b + N + 1);
	N = unique(b + 1, b + N + 1) - b - 1;
	for(int i = 1; i <= n; ++i) //离散化x坐标
		a[i].x = lower_bound(b + 1, b + N + 1, a[i].x) - b;
	sort(a + 1, a + n + 1, cmpx);
	for(int i = 1, j; i <= n; i = j) {
		int L = a[i].y, R;
		for(j = i; j <= n && a[j].x == a[i].x; ++j)
			R = a[j].y;
		if(R-L > 1) { //差分
			++tot, q[tot].x = a[i].x, q[tot].y = L+1, q[tot].v = 1;
			++tot, q[tot].x = a[i].x, q[tot].y = R, q[tot].v = -1;
		}
	}
	sort(a + 1, a + n + 1, cmpy);
	sort(q + 1, q + tot + 1);
	int cur = 1, ans = n;
	for(int i = 1, j; i <= n; i = j) {
		while(cur <= tot && q[cur].y <= a[i].y)
			upd(q[cur].x, q[cur].v), ++cur;
			for(j = i+1; j <= n && a[j].y == a[i].y; ++j)
				if(a[j-1].x+1 <= a[j].x-1)
					ans += qsum(a[j].x-1) - qsum(a[j-1].x); //差分
				//注意这里不能只用最靠左和最靠右的点形成的线段在BIT里查找
				//因为这段里面可能有最初就存在的黑点
	}
	printf("%d\n", ans);
}