bzoj4418 [Shoi2013]扇形面積並 掃描線+二分+樹狀數組

Description

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給定N個同心的扇形，求有多少面積，被至少K個扇形所覆蓋。

對於100%的數據，1≤n≤10⁵, 1≤m≤10⁶,1≤k≤5000,1≤ri≤10⁵,-m≤a1,a2≤m

Solution

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我們把圓心拉成直線，然後就變成了矩形覆蓋。掃描線做就可以了
由於一定是扇形，因此覆蓋次數一定是單調的。線段樹維護差分數組可以在線段樹上二分做到一個log，二分+樹狀數組也可以跑得很快

一開始寫線段樹被卡了。。

Code

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#include <stdio.h>
#include <string.h>
#include <algorithm>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)
#define lowbit(x) (x&-x)

typedef long long LL;
const int N=200005;

struct Q {
	int x,r;
} q[N*4];

LL s[N<<2];
int n,m,k,maxR;

int read() {
	int x=0,v=1; char ch=getchar();
	for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):(v),ch=getchar());
	for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
	return x*v;
}

void add(int x,int v) {
	for (;x<=maxR;x+=lowbit(x)) s[x]+=v;
}

int get(int x) {
	int res=0;
	for (;x;x-=lowbit(x)) res+=s[x];
	return res;
}

bool cmp(Q a,Q b) {
	return a.x<b.x;
}

int calc() {
	int l=1,r=maxR;
	for (;l<=r;) {
		int mid=(l+r)>>1;
		if (get(mid)>=k) l=mid+1;
		else r=mid-1;
	}
	return l-1;
}

int main(void) {
	freopen("xiaoqiao.in","r",stdin);
	freopen("xiaoqiao.out","w",stdout);
	n=read(),m=read(),k=read();
	int cnt=0;
	rep(i,1,n) {
		int r=read(),st=read(),ed=read();
		maxR=std:: max(maxR,r);
		st+=m; ed+=m;
		if (st>ed) {
			q[++cnt]=(Q) {st,r};
			q[++cnt]=(Q) {m*2,-r};
			q[++cnt]=(Q) {0,r};
			q[++cnt]=(Q) {ed,-r};
		} else {
			q[++cnt]=(Q) {st,r};
			q[++cnt]=(Q) {ed,-r};
		}
	}
	std:: sort(q+1,q+cnt+1,cmp);
	LL ans=0;
	for (int i=0,j=1;i<=m*2;i++) {
		for (int res,r;j<=cnt&&q[j].x<=i;j++) {
			res=calc();
			ans+=1LL*res*res*(q[j].x-q[j-1].x);
			r=abs(q[j].r);
			add(1,q[j].r/r); add(r+1,-q[j].r/r);
		}
	}
	printf("%lld\n", ans);
	return 0;
}