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【学习】Splay

Jacinth posted @ 2016年3月17日 10:52 in Lerning with tags 学习 平衡树 Splay bzoj Learning , 636 阅读

LBN好神啊~~暑假里讲了一个晚上的平衡树。。。虽然苯宝宝坚信自己写过一次BZOJ1588,然而程序丢失得一干二净。。。窝只能重新来惹

坑坑坑

No. 算法 状态
3224 Splay/Treap Accepted
1269 Splay/STL Accepted
1058 Splay/STL Accepted
1507 Splay/STL Accepted
1588 Splay Accepted
1208 Splay Accepted
1503 Splay Accepted
1861 Splay Accepted

BZOJ-1588 Splay模板题

[HNOI2002]营业额统计

Description
营业额统计 Tiger最近被公司升任为营业部经理,他上任后接受公司交给的第一项任务便是统计并分析公司成立以来的营业情况。 Tiger拿出了公司的账本,账本上记录了公司成立以来每天的营业额。分析营业情况是一项相当复杂的工作。由于节假日,大减价或者是其他情况的时候,营业额会出现一定的波动,当然一定的波动是能够接受的,但是在某些时候营业额突变得很高或是很低,这就证明公司此时的经营状况出现了问题。经济管理学上定义了一种最小波动值来衡量这种情况: 该天的最小波动值 当最小波动值越大时,就说明营业情况越不稳定。 而分析整个公司的从成立到现在营业情况是否稳定,只需要把每一天的最小波动值加起来就可以了。你的任务就是编写一个程序帮助Tiger来计算这一个值。 第一天的最小波动值为第一天的营业额。
Input 第一行为正整数 ,表示该公司从成立一直到现在的天数,接下来的n行每行有一个整数(有可能有负数) ,表示第i天公司的营业额。
Output 输出文件仅有一个正整数,即Sigma(每天最小的波动值) 。结果小于2^31 。
Sample Input
6
5
1
2
5
4
6
Sample Output
12
Hint
结果说明:5+|1-5|+|2-1|+|5-5|+|4-5|+|6-5|=5+4+1+0+1+1=12
数据少一行用0补足
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <algorithm>
#define NN 50010
#define INF 0x3f3f3f3f
using namespace std;

int root=0,n,ans=0,tree[NN][3],fa[NN],num[NN],tmp1,tmp2,size;

inline void rotate(int xx,int &k)
{
	int yy=fa[xx],zz=fa[yy],l,r;
	if (tree[yy][0]==xx) l=0; else l=1; r=l^1;
	if (yy==k) k=xx;
	else
	{	if (tree[zz][0]==yy) tree[zz][0]=xx; else tree[zz][1]=xx;}
	fa[xx]=zz, fa[yy]=xx, fa[tree[xx][r]]=yy;
	tree[yy][l]=tree[xx][r]; tree[xx][r]=yy;
}

inline void Splay(int xx,int &k)
{
	int yy,zz;
	while (xx!=k)
	{
		yy=fa[xx], zz=fa[yy];
		if (yy!=k)
		{
			if ((tree[yy][0]==xx)^(tree[zz][0]==yy)) rotate(xx,k);
			else rotate(yy,k);
		}
		rotate(xx,k);
	}
}

inline void Abefore(int &k,int xx)
{
	if (k==0) return;
	if (xx>=num[k]) tmp1=num[k], Abefore(tree[k][1],xx);
	else Abefore(tree[k][0],xx);
}
inline void Aafter(int &k,int xx)
{
	if (k==0) return;
	if (num[k]>=xx) tmp2=num[k], Aafter(tree[k][0],xx);
	else Aafter(tree[k][1],xx);
}

inline void Insert(int &k,int xx,int la)
{
	if (k==0)
	{size++, k=size, num[k]=xx, fa[k]=la, Splay(k,root); return;}
	if (xx<num[k]) Insert(tree[k][0],xx,k);
	else Insert(tree[k][1],xx,k);	
}

int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;++i)
	{
		int x;
		if (scanf("%d",&x)==EOF) x=0;
		tmp1=-INF, tmp2=INF;
		Abefore(root,x), Aafter(root,x);
		if (i==1) ans+=x;
		else ans+=min(x-tmp1,tmp2-x);
		Insert(root,x,0);
	}
	printf("%d\n",ans);
	return 0;
}

BZOJ-3224

Tyvj 1728 普通平衡树

Description
您需要写一种数据结构(可参考题目标题),来维护一些数,其中需要提供以下操作:
1. 插入x数
2. 删除x数(若有多个相同的数,因只删除一个)
3. 查询x数的排名(若有多个相同的数,因输出最小的排名)
4. 查询排名为x的数
5. 求x的前驱(前驱定义为小于x,且最大的数)
6. 求x的后继(后继定义为大于x,且最小的数)
Input 第一行为n,表示操作的个数,下面n行每行有两个数opt和x,opt表示操作的序号(1<=opt<=6)
Output
对于操作3,4,5,6每行输出一个数,表示对应答案
Sample Input
10
1 106465
4 1
1 317721
1 460929
1 644985
1 84185
1 89851
6 81968
1 492737
5 493598
Sample Output
106465
84185
492737
Hint
1.n的数据范围:n<=100000
2.每个数的数据范围:[-1e7,1e7]
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#define INF 0x3f3f3f3f
#define NN 100100
using namespace std;
int t[NN][2],num[NN],sum[NN],fa[NN],cnt[NN],root=0,tot=0,n;
inline void rotate(int k)
{
	int tmp=fa[k],y=fa[tmp],x=(t[tmp][1]==k);
	fa[t[tmp][x]=t[k][x^1]]=tmp;
	fa[t[k][x^1]=tmp]=k;
	fa[k]=y;
	if (y) t[y][t[y][1]==tmp]=k;
	sum[tmp]=sum[t[tmp][0]]+sum[t[tmp][1]]+cnt[tmp];
}

inline void Splay(int xx,int k)
{
	for (int i;((i=fa[xx])!=k);rotate(xx))
		if (fa[i]!=k) rotate((t[i][0]==xx)==(t[fa[i]][0]==i)?i:xx);
	if (k==0) root=xx;
	sum[xx]=sum[t[xx][0]]+sum[t[xx][1]]+cnt[xx];
}
inline void Insert(int k,int xx)
{
	if (k==0) 
	{ root=k=++tot, num[k]=xx, fa[k]=t[k][0]=t[k][1]=0, sum[k]=cnt[k]=1; return;}
	while (k)
	{
		sum[k]++;
		if (xx==num[k]) { cnt[k]++; return;	}
		int &y=t[k][xx>=num[k]];
		if (y==0)
		{y=++tot, num[y]=xx,fa[y]=k,t[y][0]=t[y][1]=0,sum[y]=cnt[y]=1;k=y; break;}
		k=y;
	}
	Splay(k,0);
}
inline void Del(int k,int xx)
{
	while (k)
	{
		sum[k]--;
		if (num[k]==xx)	{cnt[k]--; break;}
		int &y=t[k][xx>=num[k]];
		k=y;
	}
	if (cnt[k]!=0) return;
	Splay(k,0);
	if (t[k][0]==0) {root=t[k][1]; fa[root]=0; return;}
	if (t[k][1]==0) {root=t[k][0]; fa[root]=0; return;}
	k=t[root][0];
	while (k)
	{
		int &y=t[k][1];
		if (y==0) break;
		k=y;
	}
	Splay(k,root);
	t[k][1]=t[root][1], fa[t[k][1]]=k, sum[k]=sum[root]-1, fa[k]=0, root=k;
	return;
}
inline int rank(int k,int xx)
{
	int tmp=0, ans;
	while (k)
	{
		if (xx>num[k]) tmp+=sum[t[k][0]]+cnt[k], k=t[k][1];
		else
		{
			if (xx==num[k]) return tmp+sum[t[k][0]]+1;
			k=t[k][0];
		}
	}
	return tmp;
}
inline int kth(int k,int xx)
{
	while (k)
	{
		if (xx>=sum[t[k][0]]+1 && xx<=sum[t[k][0]]+cnt[k]) return num[k];
		if (xx>sum[t[k][0]]+cnt[k]) { xx-=sum[t[k][0]]+cnt[k]; k=t[k][1];}
		else k=t[k][0];
	}
}
inline int pre(int k,int xx)
{
	int ans=0;
	while (k)
	{	if (num[k]<xx) ans=num[k], k=t[k][1]; else k=t[k][0];}
	return ans;
}
inline int suc(int k,int xx)
{
	int ans=0;
	while (k)
	{	if (num[k]>xx) ans=num[k], k=t[k][0]; else k=t[k][1];}
	return ans;
}
int main()
{
	scanf("%d",&n);
	while (n--)
	{
		int type,x;
		scanf("%d%d",&type,&x);
		if (type==1) Insert(root,x);
		else if (type==2) Del(root,x);
		else if (type==3) printf("%d\n",rank(root,x));
		else if (type==4) printf("%d\n",kth(root,x));
		else if (type==5) printf("%d\n",pre(root,x));
		else if (type==6) printf("%d\n",suc(root,x));
	}
	return 0;
}

BZOJ-1208

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define LL long long 
#define NN 80010
#define mod 1000000
using namespace std;
int n,size,root,kind,tmp1,tmp2,tree[NN][2],num[NN],fa[NN];
LL ans;

void rotate(int x,int &k)
{
    int y=fa[x],z=fa[y],l,r;
    if(tree[y][0]==x)l=0;else l=1;r=l^1;
    if(y==k)k=x;
    else{if(tree[z][0]==y)tree[z][0]=x;else tree[z][1]=x;}
    fa[x]=z;fa[y]=x;fa[tree[x][r]]=y;
    tree[y][l]=tree[x][r];tree[x][r]=y;
}
inline void Splay(int xx,int &k)
{
	int y,z;
	while (xx!=k)
	{
		y=fa[xx], z=fa[y];
		if (y!=k)
		{
			if ((tree[y][0]==xx)^(tree[z][0]==y)) rotate(xx,k);	else rotate(y,k);
		}
		rotate(xx,k);
	}
}
inline void Del(int xx)
{
	Splay(xx,root);
	if (tree[xx][0]*tree[xx][1]==0) root=tree[xx][0]+tree[xx][1];
	else
	{
		int k=tree[xx][1];
		while (tree[k][0]) k=tree[k][0];
		tree[k][0]=tree[xx][0], fa[tree[xx][0]]=k;
		root=tree[xx][1];
	}
	fa[root]=0;
}
inline void Insert(int &k,int xx,int la)
{
	if (k==0) 
	{ k=++size, num[k]=xx, fa[k]=la, Splay(k,root); return;}
	if (xx<num[k]) Insert(tree[k][0],xx,k); else Insert(tree[k][1],xx,k);
}
inline void Aafter(int k,int xx)
{
	if (k==0) return;
	if (num[k]>=xx) tmp2=k, Aafter(tree[k][0],xx);
	else Aafter(tree[k][1],xx);
}
inline void Abefore(int k,int xx)
{
	if (k==0) return;
	if (num[k]<=xx) tmp1=k, Abefore(tree[k][1],xx);
	else Abefore(tree[k][0],xx);
}
int main()
{
	scanf("%d",&n);
	int x,y;
	for (int i=1;i<=n;++i)
	{
		scanf("%d%d",&x,&y);
		if (!root) kind=x, Insert(root,y,0);
		else if (kind==x) Insert(root,y,0);
		else
		{
			tmp1=tmp2=-1;
			Abefore(root,y), Aafter(root,y);
			if (tmp1==-1) ans+=num[tmp2]-y, ans%=mod, Del(tmp2);
			else if (tmp2==-1) ans+=y-num[tmp1], ans%=mod, Del(tmp1);
			else
				if (y-num[tmp1]>num[tmp2]-y) ans=(ans+num[tmp2]-y)%mod, Del(tmp2);
				else ans=(ans+y-num[tmp1])%mod, Del(tmp1);
		}
	}
	printf("%lld\n",ans);
	return 0;
}

BZOJ-1507

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#define NN 2010
using namespace std;

int n,m,a[NN][NN],f[NN][NN],l[NN][NN],r[NN][NN],h[NN][NN];

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;++i)
		for (int j=1;j<=m;++j) scanf("%d",&a[i][j]);
	int ans1=1;
	for (int i=1;i<=n;++i)
		for (int j=1;j<=m;++j)
		{
			if (i==1 || j==1) f[i][j]=1;
			else if (a[i][j]==a[i-1][j] || a[i][j]==a[i][j-1]) f[i][j]=1;
			else f[i][j]=min(f[i-1][j-1],min(f[i][j-1],f[i-1][j]))+1;
			ans1=max(ans1,f[i][j]);
		}
	printf("%d\n",ans1*ans1);
	for (int i=1;i<=n;++i)
		for (int j=1;j<=m;++j)
		{
			if (i==1) h[i][j]=1;
			else if (a[i][j]==a[i-1][j]) h[i][j]=1;
			else h[i][j]=h[i-1][j]+1;
		}
	for (int i=1;i<=n;++i)
		for (int j=1;j<=m;++j)
		{
			if (j==1) l[i][j]=j;
			else if (a[i][j]==a[i][j-1]) l[i][j]=j;
			else if (h[i][j-1]<h[i][j]) l[i][j]=j;
			else
			{
				l[i][j]=l[i][j-1];
				while (l[i][j]>1 && h[i][l[i][j]-1]>=h[i][j] && a[i][l[i][j]]!=a[i][l[i][j]-1])
					l[i][j]=l[i][l[i][j]-1];
			}
		}
	for (int i=1;i<=n;++i)
		for (int j=m;j>=1;--j)
		{
			if (j==m) r[i][j]=j;
			else if (a[i][j]==a[i][j+1]) r[i][j]=j;
			else if (h[i][j+1]<h[i][j]) r[i][j]=j;
			else
			{
				r[i][j]=r[i][j+1];
				while (r[i][j]<m && h[i][r[i][j]+1]>=h[i][j] && a[i][r[i][j]]!=a[i][r[i][j]+1])
					r[i][j]=r[i][r[i][j]+1];
			}
		}
	int ans2=0;
	for (int i=1;i<=n;++i)
		for (int j=1;j<=m;++j)
			ans2=max(ans2,h[i][j]*(r[i][j]-l[i][j]+1));
	printf("%d\n",ans2);
	return 0;
}

BZOJ-1861

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define NN 80050
#define LL long long 
#define INF 0x3f3f3f3f
using namespace std;

int n,m,size[NN],fa[NN],tr[NN][2],pos[NN],a[NN],v[NN],root=0;
void update(int k)
{
	size[k]=size[tr[k][0]]+size[tr[k][1]]+1;
}

void rotate(int x,int &k)
{
	int y=fa[x],z=fa[y],l=(tr[y][0]!=x),r=l^1;
	if (y==k) k=x; 
	else if (tr[z][0]==y) tr[z][0]=x; else tr[z][1]=x;
	fa[x]=z, fa[y]=x, fa[tr[x][r]]=y;
	tr[y][l]=tr[x][r], tr[x][r]=y;
	update(y), update(x);
}
void Splay(int x,int &k)
{
	while (x!=k)
	{
		int y=fa[x],z=fa[y];
		if (y!=k)
		{
			if ((tr[y][0]==x)^(tr[z][0]==y)) rotate(x,k); else rotate(y,k);
		}
		rotate(x,k);
	}
}
void Build(int l,int r,int la)
{
	if (l>r)  return;
	int now=l, last=la;
	if (l==r)
	{
		v[now]=a[l], size[now]=1, fa[now]=last;
		if (l<la) tr[last][0]=now; else tr[last][1]=now;
		return;
	}
	int mid=(l+r)>>1; now=mid;
	Build(l,mid-1,mid), Build(mid+1,r,mid);
	v[now]=a[mid], fa[now]=last, update(now);
	if (mid<la) tr[last][0]=now; else tr[last][1]=now;
}
int Find(int k,int rank)
{
	int l=tr[k][0], r=tr[k][1];
	if (size[l]+1==rank) return k;
	else if (size[l]>=rank) return Find(l,rank);
	else return Find(r,rank-size[l]-1);
}
void Del(int k)
{
	int x=Find(root,k-1),y=Find(root,k+1),z;
	Splay(x,root), Splay(y,tr[x][1]);
	z=tr[y][0], tr[y][0]=0, fa[z]=size[z]=0;
	update(y), update(x);
}
void move(int k,int v)
{
	int x,y,z=pos[k],rank;
	Splay(z,root); rank=size[tr[z][0]]+1;
	Del(rank);
	if (v==INF) x=Find(root,n), y=Find(root,n+1);
	else if (v==-INF) x=Find(root,1), y=Find(root,2);
	else x=Find(root,rank+v-1), y=Find(root,rank+v);
	Splay(x,root), Splay(y,tr[x][1]);
	size[z]=1, fa[z]=y, tr[y][0]=z;
	update(y), update(x);
}
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=2;i<=n+1;++i) scanf("%d",&a[i]),pos[a[i]]=i;
	Build(1,n+2,0);
	root=(n+3)>>1;
	char ch[10]; int x,y;
	for (int i=1;i<=m;++i)
	{
		scanf("%s",ch); scanf("%d",&x);
		if (ch[0]=='T') move(x,-INF);
		else if (ch[0]=='B') move(x,INF);
		else if (ch[0]=='I') scanf("%d",&y), move(x,y);
		else if (ch[0]=='A') Splay(pos[x],root), printf("%d\n",size[tr[pos[x]][0]]-1);
		else if (ch[0]=='Q') printf("%d\n",v[Find(root,x+1)]);
	}
	return 0;
}

BZOJ-1269

发现rope方法的长度和Splay的长度真是天壤之别

于是就开始学习偷懒技巧QAQ

//Rope
#include <cstdio>
#include <iostream>
#include <ext/rope>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;
using namespace __gnu_cxx;

inline int read()
{
    int x=0;char ch=getchar();
    while(ch>'9'||ch<'0')ch=getchar();
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x;
}

#define NN 2000010

int n,now,len,x;
rope <char> a,b,tmp;
char ch[NN],chh[NN];

int main()
{
	n = read();
	while (n--)
	{
		scanf("%s",ch);
		switch (ch[0])
		{
			case 'M': now = read(); break;
			case 'P': now --; break;
			case 'N': now ++; break;
			case 'G': printf("%c\n",a[now]); break;
			case 'I':
				x = read(), len = a.length();
				for (int i=0;i<x;++i)
				{
					ch[i] = getchar();
					while (ch[i] == '\n') ch[i] = getchar();
					chh[x-i-1] = ch[i];
				}
				chh[x] = ch[x] = 0;
				a.insert(now,ch);
				b.insert(len-now,chh);
				break;
			case 'D':
				x = read(), len = a.length();
				a.erase(now,x);
				b.erase(len-now-x,x);
				break;
			case 'R':
				x = read(), len = a.length();
				tmp = a.substr(now,x);
				a = a.substr(0,now) + b.substr(len-now-x,x) + a.substr(now+x,len-now-x);
				b = b.substr(0,len-now-x) + tmp + b.substr(len-now,now);
				break;
		}
	}
	return 0;
}

BZOJ-1058

STL大法好啊QAQ

披着Splay外衣的STL

恩,窝还是把塔扔进了这里QAQ

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <set>
#include <map>
#include <queue>
using namespace std;

#define INF 0x3f3f3f3f
#define NN 500005

char ch[20];
int n,m,st[NN],ed[NN],p,x,tmp;
map <int,int> mp;
priority_queue <int, vector<int>,greater<int> > que;
multiset <int> a,b;

inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

inline void P(int xx)
{
	int l=*--b.lower_bound(xx), r = *b.lower_bound(xx);
	que.push(min(xx -l, r -xx));
	b.insert(xx);
}

inline void Ins(int xx)
{
	mp[xx] ++;
	if (mp[xx] == 1) a.insert(xx);
}

int main()
{
	n = read(), m = read();
	b.insert(INF), b.insert(-INF);
	for (int i=1;i<=n;++i)
	{
		x = read();
		st[i] = ed[i] = x;
		P(x);
	}
	for (int i=2;i<=n;++i) Ins(abs(st[i] - st[i-1]));
	while (m--)
	{
		scanf("%s",ch);
		if (ch[0] == 'I')
		{
			p = read(), x = read();
			if (p != n)
			{
				tmp = abs(ed[p] - st[p+1]);
				mp[tmp] --;
				if (!mp[tmp]) a.erase(tmp);
			}
			Ins(abs(ed[p] - x));
			Ins(abs(x - st[p+1]));
			ed[p] = x;
			P(x);
		}
		else if (ch[4] == 'S') printf("%d\n",que.top());
		else printf("%d\n",*a.begin());
	}
	return 0;
}

BZOJ-1503

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define NN 100010
#define INF 0x3f3f3f3f
using namespace std;

int sum[NN],tr[NN][2],fa[NN],num[NN],n,minn,root,size,delta=0,le=0;

inline void update(int k) {sum[k]=sum[tr[k][0]]+sum[tr[k][1]]+1;}
inline void rotate(int x,int &k)
{
	int y=fa[x],z=fa[y],l=(tr[y][0]!=x),r=l^1;
	if (y==k) k=x;
	else if (tr[z][0]==y) tr[z][0]=x; else tr[z][1]=x;
	fa[x]=z, fa[y]=x, fa[tr[x][r]]=y;
	tr[y][l]=tr[x][r], tr[x][r]=y;
	update(y), update(x);
}
inline void Splay(int x,int &k)
{
	int y,z;
	while (x!=k)
	{
		y=fa[x], z=fa[y];
		if (y!=k)
			if ((tr[y][0]==x)^(tr[z][0]==y)) rotate(x,k); else rotate(y,k);
		rotate(x,k);
	}
}
inline int Dec(int &k,int xx)
{
	if (!k) return 0;
	int x;
	if (num[k]+delta<minn)
	{
		x=Dec(tr[k][1],k)+sum[tr[k][0]]+1;
		sum[tr[k][1]]=sum[k]-x;
		k=tr[k][1], fa[k]=xx;
	}
	else x=Dec(tr[k][0],k), sum[k]-=x;
	return x;
}
inline int Find(int k,int rank)
{
	int l=tr[k][0], r=tr[k][1];
	if (rank<=sum[r]) return Find(r,rank);
	if (rank==sum[r]+1)  return k;
	return Find(l,rank-sum[r]-1);
}
inline void Insert(int &k,int xx,int la)
{
	if (k==0) {k=++size, num[k]=xx, fa[k]=la, sum[k]=1, Splay(k,root); return;}
	if (xx<num[k]) Insert(tr[k][0],xx,k); else Insert(tr[k][1],xx,k);
}
int main()
{
	scanf("%d%d",&n,&minn);
	char ch[10];
	int x;
	for (int i=1;i<=n;++i)
	{
		scanf("%s",ch); scanf("%d",&x);
		switch (ch[0])
		{
			case'I':if (x>=minn) Insert(root,x-delta,0); break;
			case'A':delta+=x; break;
			case'S':delta-=x; le+=Dec(root,0); break;
			case'F':printf("%d\n",x<=sum[root]?num[Find(root,x)]+delta:-1); break;
		}
	}
	printf("%d\n",le);
	return 0;
}
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NCERT Civics Questio 说:
2022年9月26日 02:14

Civics is also a part of Social Studies for all Primary or Junior School students studying in Hindi medium, English medium and Urdu medium schools located in all locations of the Country. By understanding Civics subjects students can get aware of their life as good civilians and Civics also a inclusion of voting, NCERT Civics Question Paper Class 6 volunteering, participating in group activities, and community gardening which make the students good citizens.Civics is also a part of Social Studies for all Primary or Junior School students studying in Hindi medium, English medium and Urdu medium schools located in all locations of the Country. By understanding Civics subjects students can get aware of their life as good civilians and Civics also a inclusion of voting.


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