[乱搞 树状数组] BZOJ 4548 小奇的糖果 && BZOJ 3658 Jabberwocky

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跟悬线法有点像

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#define cl(x) memset(x,0,sizeof(x))
using namespace std;

inline char nc()
{
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(int &x)
{
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=100005;

namespace BIT{
	#define lowbit(x) ((x)&-(x))
	int maxn,c[N];
	inline void init(int n){
		maxn=n; cl(c);
	}
	inline void add(int x,int r){
		for (int i=x;i<=maxn;i+=lowbit(i))
			c[i]+=r;
	}
	inline int sum(int x){
		int ret=0;
		for (int i=x;i;i-=lowbit(i))
			ret+=c[i];
		return ret;
	}
	inline int sum(int l,int r){
		if (r<l) return 0;
		return sum(r)-sum(l-1);
	}
}

struct abcd{
	int x,y,w;
	int idx;
	int rx;
}a[N];

bool cmpx(abcd A,abcd B){
	return A.x<B.x;
}

bool cmpy(abcd A,abcd B){
	return A.y<B.y;
}

int sx[N],icnt;

inline int Bin(int x){
	return lower_bound(sx+1,sx+icnt+1,x)-sx;
}

int n,K,ans;
int pos[N],last[N];
int l[N],r[N];

inline void Solve()
{
	using namespace BIT;
	int idx,lst;
	init(icnt); cl(last);
	for (int i=1;i<=n;i++) add(a[i].x,1);
	for (int i=1;i<=n;i++) l[i]=0,r[i]=n+1;
	sort(a+1,a+n+1,cmpx);
	pos[0]=0; pos[n+1]=icnt+1;
	for (int i=1;i<=n;i++){
		a[i].rx=i; pos[i]=a[i].x;
		lst=last[a[i].w]; last[a[i].w]=i;
		l[i]=lst; if (lst) r[lst]=i;
		ans=max(ans,sum(a[lst].x+1,a[i].x-1));
	}
	for (int i=1;i<=K;i++) 
		ans=max(ans,sum(a[last[i]].x+1,icnt));
	sort(a+1,a+n+1,cmpy);
	for (int i=1,j=1;i<=n;i++)
	{
		while (j<=n && a[i].y==a[j].y) add(a[j++].x,-1);
		ans=max(ans,sum(pos[l[a[i].rx]]+1,pos[r[a[i].rx]]-1));
		if (r[a[i].rx]) l[r[a[i].rx]]=l[a[i].rx]; 
		if (l[a[i].rx]) r[l[a[i].rx]]=r[a[i].rx];
		r[a[i].rx]=l[a[i].rx]=0;
	}
}

int main()
{
	int Q; 
	freopen("t.in","r",stdin);
	freopen("t.out","w",stdout);
	read(Q);
	while (Q--)
	{
		icnt=0; ans=0;
		read(n); read(K);
		for (int i=1;i<=n;i++)
			read(a[i].x),read(a[i].y),read(a[i].w),sx[++icnt]=a[i].x;
		sort(sx+1,sx+icnt+1);
		icnt=unique(sx+1,sx+icnt+1)-sx-1;
		for (int i=1;i<=n;i++)
			a[i].x=Bin(a[i].x),pos[i]=a[i].x;
		Solve();
		for (int i=1;i<=n;i++) a[i].y=-a[i].y;
		Solve();
		printf("%d\n",ans);
	}
	return 0;
}

[乱搞 树状数组] BZOJ 4548 小奇的糖果 && BZOJ 3658 Jabberwocky

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原文地址：http://blog.csdn.net/u014609452/article/details/51345187