2018HDU多校训练-3-Problem F. Grab The Tree
In this game, Little Q needs to grab some vertices on the tree. He can select any number of vertices to grab, but he is not allowed to grab both vertices that are adjacent on the tree. That is, if there is an edge between x
and y
, he can't grab both x
and y
. After Q's move, Little T will grab all of the rest vertices. So when the game finishes, every vertex will be occupied by either Q or T.
The final score of each player is the bitwise XOR sum of his choosen vertices' value. The one who has the higher score will win the game. It is also possible for the game to end in a draw. Assume they all will play optimally, please write a program to predict the result.
, denoting the number of test cases.
In each test case, there is one integer n(1≤n≤100000)
in the first line, denoting the number of vertices.
In the next line, there are n
integers w1,w2,...,wn(1≤wi≤109)
, denoting the value of each vertex.
For the next n−1
lines, each line contains two integers u
and v
, denoting a bidirectional edge between vertex u
and v
.
3
2 2 2
1 2
1 3
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int w[maxn],u[maxn],v[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); int T,n;
cin>>T;
while(T--)
{
cin>>n;
bool flag=false;
for(int i=1;i<=n;i++) cin>>w[i];
for(int i=1;i<n;i++) cin>>u[i]>>v[i];
for(int i=0;i<32;i++)
{
int cnt=0;
for(int i=1;i<=n;i++)
{
if(w[i]&1) cnt++;
w[i]>>=1;
}
if(cnt & 1)
{
cout<<"Q"<<endl;
flag=true;
break;
}
}
if(!flag) cout<<"D"<<endl;
} return 0;
}
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