Test room: decoration (tree-like pressure +)

topic:

 

 

 

 analysis:

This question lies in the difficulty:

Every time the transfer, to record a lot of points to be flipped, but also to determine who their next fa yes.

So to change my ideas: the transition from state to terminate 0 state (initial state), backwards to.

Suppose now to the penultimate i seconds, then flipped a light from inside the i ~ ans time he is been going to jump up and affect other lamps, jump time i, it may be pretreated at: FA [i] [ j] i represents the lamp after step j can be varied to jump state

When each transfer, just enumerate lamp, XOR array can quickly know the fa penultimate second state i + 1.

An update to the initial state 0, direct output now is the inverse of the first few seconds can be.

Because even recording time, so dp to open two-dimensional.

Note: If a point has been passed to the root node, then the state will not become a! !

 

#include<bits/stdc++.h>
using namespace std;
#define N 18
#define ri register int
int ff[N],fa[N][N],dp[N][1<<N],n,goal=0,x;
int main()
{
    freopen("decoration.in","r",stdin);
    freopen("decoration.out","w",stdout);
    scanf("%d",&n);
    ff[1]=0;
    for(ri i=2;i<=n;++i) scanf("%d",&x),ff[i]=x;
    for(ri i=1;i<=n;++i) scanf("%d",&x),goal<<=1,goal+=x;
    for(ri i=1;i<=n;++i){
        fa[i][0]=(1<<(n-i)); 
        int pre=ff[i];
        for(ri j=1;j<=n;++j){
            if(pre) fa[i][j]=fa[i][j-1]^(1<<(n-pre));// If the root node has been reached 1, and would not jump up 
            the else FA [I] [J] = FA [I] [J- 1 ]; 
            pre = FF [pre]; 
        } 
    } 
    DP [ 0 ] [Goal ] = . 1 ;
     for (RI T = 0 ; T <= n-; ++ T) {
         IF (DP [T] [ 0 ]) {the printf ( " % D \ n- " , T); return  0 ;}
         for ( J = RI 0 ; J <( . 1 << n-); ++ J)
         IF (DP [T] [J]) { 
            DP [T + . 1 ] [J] =1;
            for(ri k=1;k<=n;++k)
            dp[t+1][j^(fa[k][t])]=1;
        }
    }
    return 0;
}
/*
7
1 1 2 2 3 3
0 1 1 1 0 0 1
3

7
1 1 2 2 3 3
1 1 1 1 0 0 1
3

8
1 1 2 4 2 4 6
0 0 1 0 1 1 0 1


3 
1 1
0 1 1
*/

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