[HDOJ 4866] Shooting [可持久化线段树]

在一个平面直角坐标系上，有一些靶子，靶子可以视为平行于x轴的线段，有一个人站在某个位置打靶，打靶的方向永远是y轴正方向。他每次可以打穿前k-1个靶子，所以共击中前k个靶子，击中一个靶子可以得到那个靶子与x轴的距离的分数。靶子被击中后仍然存在。问这个人每次打靶得到的分数。如果之前打靶的分数超过了p，那么下次打靶得到的分数会乘2。

数据范围：x轴大小为10^5，y轴大小为10^7，k为每次根据上一次结果和输入数据给的一些参数计算得出。

做法：把靶子按照y轴坐标排序，用一棵可持久化线段树，按照距离依次将靶子添加到线段树上，即能打到这个靶子的区间整体加上这个靶子的分数。查询时先二分答案，判断他能打到前k个靶子是具体能打到排序后的第几个靶子，然后在对应的线段树上查询能得到的分数。

由于内存要求不够，所以又把区间修改点查询转换成了点修改区间查询。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

struct SegTreeNode {
	SegTreeNode *ls,*rs;
	long long score;
	int count;
};
SegTreeNode a[4000000];
SegTreeNode *acur;
struct PersistentSegTree {
	SegTreeNode *root[200001];
	SegTreeNode *maketree(int l,int r) {
		SegTreeNode *ans=acur++;
		ans->score=ans->count=0;
		if (l==r) {
			ans->ls=ans->rs=NULL;
		} else {
			int t=(l+r)/2;
			ans->ls=maketree(l,t);
			ans->rs=maketree(t+1,r);
		}
		return ans;
	}
	void clear(int l,int r) {
		acur=a;
		root[0]=maketree(l,r);
	}
	SegTreeNode *set(SegTreeNode *from,int l,int r,int pos,int x) {
		//printf("%d %d %d %d %d\n",l,r,ll,rr,x);
		SegTreeNode *ans=acur++;
		*ans=*from;
		if (l==r) {
			ans->score+=x;
			if (x>0) ans->count++;
			else ans->count--;
		} else {
			int t=(l+r)/2;
			if (pos<=t) {
				ans->ls=set(from->ls,l,t,pos,x);
			} else {
				ans->rs=set(from->rs,t+1,r,pos,x);
			}
			ans->score=ans->ls->score+ans->rs->score;
			ans->count=ans->ls->count+ans->rs->count;
		}
		//printf("%lld %d\n",ans->score,ans->count);
		return ans;
	}
	int getnum(SegTreeNode *from,int l,int r,int ll,int rr) {
		if (l==ll&&r==rr) return from->count;
		int t=(l+r)/2;
		if (rr<=t) {
			return getnum(from->ls,l,t,ll,rr);
		} else if (ll>t) {
			return getnum(from->rs,t+1,r,ll,rr);
		} else {
			return getnum(from->ls,l,t,ll,t)+getnum(from->rs,t+1,r,t+1,rr);
		}
	}
	long long getscore(SegTreeNode *from,int l,int r,int ll,int rr) {
		if (l==ll&&r==rr) return from->score;
		int t=(l+r)/2;
		if (rr<=t) {
			return getscore(from->ls,l,t,ll,rr);
		} else if (ll>t) {
			return getscore(from->rs,t+1,r,ll,rr);
		} else {
			return getscore(from->ls,l,t,ll,t)+getscore(from->rs,t+1,r,t+1,rr);
		}
	}
};
struct Target{
	int l,r,d;
	void read() {
		scanf("%d%d%d",&l,&r,&d);
	}
	friend bool operator < (const Target &a,const Target &b) {
		return a.d<b.d;
	}
};

int n,m,maxx,p;
Target b[100000];
PersistentSegTree c;

int main() {
	int i;
	long long pre;
	while (scanf("%d%d%d%d",&n,&m,&maxx,&p)!=EOF) {
		for (i=0;i<n;i++) {
			b[i].read();
		}
		sort(b,b+n);
		maxx++;
		c.clear(1,maxx);
		//printf("Build 0 complete\n");
		for (i=0;i<n;i++) {
			c.root[i*2+1]=c.set(c.root[i*2],1,maxx,b[i].l,b[i].d);
			c.root[i*2+2]=c.set(c.root[i*2+1],1,maxx,b[i].r+1,-b[i].d);
			//printf("Build %d complete\n",i+1);
		}
		//printf("Build complete\n");
		pre=1;
		for (i=0;i<m;i++) {
			int xx,k,aa,bb,cc;
			scanf("%d%d%d%d",&xx,&aa,&bb,&cc);
			k=(aa*pre+bb)%cc;
			int l=0,r=n;
			while (l!=r) {
				int t=(l+r)/2;
				int tmp=c.getnum(c.root[t*2],1,maxx,1,xx);
				if (tmp<k) l=t+1;
				else r=t;
			}
			//printf("%d %d\n",k,c.getnum(c.root[l],1,maxx,xx));
			long long ans=c.getscore(c.root[l*2],1,maxx,1,xx);
			if (pre>p) ans*=2;
			printf("%I64d\n",ans);
			pre=ans;
		}
		//static int cas=1;
		//cerr<<cas++<<"complete\n";
		//cerr<<acur-a<<endl;
	}
	return 0;
}

附上原来的点修改区间查询版本...内存超了30M....

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

struct SegTreeNode {
	SegTreeNode *ls,*rs;
	long long score;
	int count;
};
SegTreeNode a[8000000];
SegTreeNode *acur;
struct PersistentSegTree {
	SegTreeNode *root[100001];
	SegTreeNode *maketree(int l,int r) {
		SegTreeNode *ans=acur++;
		ans->score=ans->count=0;
		if (l==r) {
			ans->ls=ans->rs=NULL;
		} else {
			int t=(l+r)/2;
			ans->ls=maketree(l,t);
			ans->rs=maketree(t+1,r);
		}
		return ans;
	}
	void clear(int l,int r) {
		acur=a;
		root[0]=maketree(l,r);
	}
	SegTreeNode *set(SegTreeNode *from,int l,int r,int ll,int rr,int x) {
		//printf("%d %d %d %d %d\n",l,r,ll,rr,x);
		SegTreeNode *ans=acur++;
		*ans=*from;
		if (l==ll&&r==rr) {
			ans->score+=x;
			ans->count++;
		} else {
			int t=(l+r)/2;
			if (rr<=t) {
				ans->ls=set(from->ls,l,t,ll,rr,x);
			} else if (ll>t) {
				ans->rs=set(from->rs,t+1,r,ll,rr,x);
			} else {
				ans->ls=set(from->ls,l,t,ll,t,x);
				ans->rs=set(from->rs,t+1,r,t+1,rr,x);
			}
		}
		//printf("%lld %d\n",ans->score,ans->count);
		return ans;
	}
	int getnum(SegTreeNode *from,int l,int r,int x) {
		int count=0;
		while (l!=r) {
			count+=from->count;
			int t=(l+r)/2;
			if (x<=t) {
				r=t;
				from=from->ls;
			} else {
				l=t+1;
				from=from->rs;
			}
		}
		count+=from->count;
		return count;
	}
	long long getscore(SegTreeNode *from,int l,int r,int x) {
		long long score=0;
		while (l!=r) {
			score+=from->score;
			int t=(l+r)/2;
			if (x<=t) {
				r=t;
				from=from->ls;
			} else {
				l=t+1;
				from=from->rs;
			}
		};
		score+=from->score;
		return score;
	}
};
struct Target{
	int l,r,d;
	void read() {
		scanf("%d%d%d",&l,&r,&d);
	}
	friend bool operator < (const Target &a,const Target &b) {
		return a.d<b.d;
	}
};

int n,m,maxx,p;
Target b[100000];
PersistentSegTree c;

int main() {
	int i;
	long long pre;
	while (scanf("%d%d%d%d",&n,&m,&maxx,&p)!=EOF) {
		for (i=0;i<n;i++) {
			b[i].read();
		}
		sort(b,b+n);
		c.clear(1,maxx);
		for (i=0;i<n;i++) {
			//printf("-- %d %d %d\n",b[i].l,b[i].r,b[i].d);
			c.root[i+1]=c.set(c.root[i],1,maxx,b[i].l,b[i].r,b[i].d);
		}
		pre=1;
		/*
		for (i=0;i<=n;i++) {
			printf("%dth tree\n",i);
			for (int j=1;j<=maxx;j++) printf("%d ",c.getnum(c.root[i],1,maxx,j));
			printf("\n");
			for (int j=1;j<=maxx;j++) printf("%lld ",c.getscore(c.root[i],1,maxx,j));
			printf("\n");
		}
		*/
		for (i=0;i<m;i++) {
			int xx,k,aa,bb,cc;
			scanf("%d%d%d%d",&xx,&aa,&bb,&cc);
			k=(aa*pre+bb)%cc;
			int l=0,r=n;
			while (l!=r) {
				int t=(l+r)/2;
				int tmp=c.getnum(c.root[t],1,maxx,xx);
				if (tmp<k) l=t+1;
				else r=t;
			}
			//printf("%d %d\n",k,c.getnum(c.root[l],1,maxx,xx));
			long long ans=c.getscore(c.root[l],1,maxx,xx);
			if (pre>p) ans*=2;
			printf("%lld\n",ans);
			pre=ans;
		}
		//static int cas=1;
		//cerr<<cas++<<"complete\n";
		//cerr<<acur-a<<endl;
	}
	return 0;
}