CF: 368 (Div2 D) [ BITSET+DFS ]

CF: 368 (Div. 2) [ BITSET+DFS ]
///==================================================///
/// HELLO WORLD !! ///
/// IT'S ME ///
/// BISHAL GAUTAM ///
/// [ bsal.gautam16@gmail.com ] ///
///==================================================///
#include<bits/stdc++.h>
#define X first
#define Y second
#define mpp make_pair
#define nl printf("\n")
#define SZ(x) (int)(x.size())
#define pb(x) push_back(x)
#define pii pair<int,int>
#define pll pair<ll,ll>
///---------------------
#define S(a) scanf("%d",&a)
#define P(a) printf("%d",a)
#define SL(a) scanf("%lld",&a)
#define S2(a,b) scanf("%d%d",&a,&b)
#define SL2(a,b) scanf("%lld%lld",&a,&b)
///------------------------------------
#define all(v) v.begin(),v.end()
#define CLR(a) memset(a,0,sizeof(a))
#define SET(a) memset(a,-1,sizeof(a))
#define fr(i,a,n) for(int i=a;i<=n;i++)
using namespace std;
typedef long long ll;

/// Digit 0123456789012345678 ///
#define MX 100005
#define inf 2000000010
#define MD 1000000007
#define eps 1e-9
///===============================///

struct data {
int op,x,y;
data() {};
data(int o,int a,int b) {
op=o,x=a,y=b;
}
} Q[MX+2];

vector<int>G[MX+2];
bitset<1001>dp[1001],lim;
int n,m,ans[MX+2];

void Dfs(int u,int sm) {
ans[u]=sm;
for(int i=0; i<SZ(G[u]); i++) {
int v=G[ u ][ i ];
int op=Q[v].op;
int x=Q[v].x;
int y=Q[v].y;
if( op==1 ) {
if( !dp[x].test(y) ) {
dp[x][ y ]=1;
Dfs(v,sm+1);
dp[x][ y ]=0;
} else Dfs(v,sm);
}
else if( op==2 ) {
if( dp[x].test(y) ) {
dp[x][ y ]=0;
Dfs(v,sm-1);
dp[x][ y ]=1;
} else Dfs(v,sm);
}
else if( op==3 ){
int pc=dp[x].count();
dp[ x ]^=lim;
int nc=m-pc;
Dfs( v,sm-pc+nc );
dp[ x ]^=lim;
}else Dfs(v,sm);
}
}

int main() {
int q,i,j,k,x,y,z,op;
S2(n,m);
S(q);
fr(i,1,m)lim.set(i); ///Making Full SetBit
fr(i,1,q) {
S(op);
y=-1;
///Time Machine DAG
if(op<=2)S2(x,y),G[i-1].pb( i );
else S(x);
Q[i]=data(op,x,y);
if(op==3)G[ i-1 ].pb( i );
if(op==4)G[ x ].pb( i );
}
Dfs(0,0); ///Answering Queries On the basis of DFS tree
fr(i,1,q){
printf("%d\n",ans[i]);
}
return 0;
}


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