Cf1010d Mars Rover [bitwise operation]

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Mars Rover

The format is difficult to adjust.

 

Analysis:

I took an enhanced version of this question during the exam today.

In fact, it is not difficult. After establishing a tree structure, we first obtain all the initial weights and then obtain the initial values of the root node. Because only the value of one vertex is modified at a time, we only need to determine from top to bottom whether the value of each node will change the value of the root node after being modified based on the bit operation type.

Code:

　　

//It is made by HolseLee on 2nd Nov 2018//CF1010D#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int N=1e6+7;int n,dp[N];struct Node { int ls,rs,val,type; }t[N];inline int read(){    char ch=getchar(); int x=0; bool flag=false;    while( ch<‘0‘ || ch>‘9‘ ) {        if( ch==‘-‘ ) flag=true; ch=getchar(); }    while( ch>=‘0‘ && ch<=‘9‘ ) {        x=x*10+ch-‘0‘; ch=getchar(); }    return flag ? -x : x;}void dfs(int x){    if( t[x].type==5 ) return;    switch (t[x].type) {        case 1:            dfs(t[x].ls), dfs(t[x].rs);            t[x].val=t[t[x].ls].val&t[t[x].rs].val;            break;        case 2:            dfs(t[x].ls), dfs(t[x].rs);            t[x].val=t[t[x].ls].val^t[t[x].rs].val;            break;        case 3:            dfs(t[x].ls), dfs(t[x].rs);            t[x].val=t[t[x].ls].val|t[t[x].rs].val;            break;        case 4:            dfs(t[x].ls); t[x].val=(t[t[x].ls].val^1);            break;    }}void DP(int x){    if( t[x].type==5 ) return;    switch (t[x].type) {        case 1:            if( t[t[x].ls].val==1 && t[t[x].rs].val==1 ) {                dp[t[x].ls]=dp[t[x].rs]=1;                DP(t[x].ls), DP(t[x].rs);            } else if( t[t[x].ls].val==1 && t[t[x].rs].val==0 ) {                dp[t[x].rs]=1; DP(t[x].rs);            } else if( t[t[x].ls].val==0 && t[t[x].rs].val==1 ) {                dp[t[x].ls]=1; DP(t[x].ls);            }            break;        case 2:            dp[t[x].ls]=dp[t[x].rs]=1;            DP(t[x].ls), DP(t[x].rs);            break;        case 3:            if( t[t[x].ls].val==1 && t[t[x].rs].val==0 ) {                dp[t[x].ls]=1, DP(t[x].ls);            } else if( t[t[x].ls].val==0 && t[t[x].rs].val==1 ) {                dp[t[x].rs]=1, DP(t[x].rs);            } else if( t[t[x].ls].val==0 && t[t[x].rs].val==0 ) {                dp[t[x].ls]=dp[t[x].rs]=1;                DP(t[x].ls), DP(t[x].rs);            }            break;        case 4:            dp[t[x].ls]=1, DP(t[x].ls);            break;    }}int main(){    n=read(); char s[10];    for(int i=1; i<=n; ++i) {        scanf("%s",s);        switch (s[0]) {            case ‘A‘: t[i].type=1, t[i].ls=read(), t[i].rs=read(); break;            case ‘X‘: t[i].type=2, t[i].ls=read(), t[i].rs=read(); break;            case ‘O‘: t[i].type=3, t[i].ls=read(), t[i].rs=read(); break;            case ‘N‘: t[i].type=4, t[i].ls=read(); break;            case ‘I‘: t[i].type=5, t[i].val=read(); break;        }    }    dfs(1); DP(1);    for(int i=1; i<=n; ++i) {        if( t[i].type!=5 ) continue;        if( dp[i] ) putchar((t[1].val^1)+‘0‘);        else putchar(t[1].val+‘0‘);    }    puts("");    return 0;}

 

Cf1010d Mars Rover [bitwise operation]