Codeforces Round #316 (Div. 2) D. Tree Requests
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Codeforces Round #316 (Div. 2) D. Tree Requests
题意:
给一棵树,每个节点上都唯一对应一个单词,给M个询问,问节点u的子树中,深度为h的节点上所有的单词以任意次序组合起来能否构成回文串?
思路:
首先呢,我们必须得处理出内节点所在的深度,而且得知道对应深度下有哪些节点存于vector数组H[d]中,但是问题就是询问仅涉及到节点为u的子树的哪些节点,所以满足条件的节点应该为数组中的某一段,那么如何去找这一段呢? 利用
dfs 序+二分查找,就能得到满足条件的节点所对应的区间,剩下的问题就是,如何求是否为回文串,因为次序任意,所以只要出现次数为奇数的单词的个数不超过2就行了,这个肯定是需要去预处理出整个区间的啦,方法就是利用位运算 + 异或 解决!
代码如下:
#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#include <iostream>#include <cmath>#include <map>#include <vector>#include <set>#include <string>#define PB push_back#define FT first#define SD second#define MP make_pair#define INF (0x3f3f3f3f)*2using namespace std;typedef long long LL;typedef pair<int,int> P;const int maxn=5+1e6,MOD=7+1e9;vector<int> G[maxn]; vector< pair<int,int> > H[maxn];char ss[maxn];int A[30],ord,in[maxn],out[maxn];void dfs(int u,int d){ in[u] = ++ord; H[d].PB(MP(ord,H[d].back().SD ^ A[ss[u]-'a'])); for(int i=0;i<G[u].size();i++){ dfs(G[u][i],d+1); } out[u] = ++ord;}int main(){ //freopen("E:/ACM_code/in.txt","r",stdin); //freopen("out.txt","w",stdout); int n,m; scanf("%d%d",&n,&m); for(int i=2;i<=n;i++){ int x; scanf("%d",&x); G[x].PB(i); } scanf("%s",ss+1); for(int i=1;i<=n;i++) H[i].resize(1); for(int i=0;i<=30;i++) A[i]= 1 << i; ord=0; dfs(1,1); //for(int i=1;i<=n;i++) cout<<i<<" "<<in[i]<<" "<<out[i]<<endl; while(m--){ int d,v; scanf("%d%d",&v,&d); int l=lower_bound(H[d].begin(),H[d].end(),MP(in[v],-1)) - H[d].begin() -1; int r=lower_bound(H[d].begin(),H[d].end(),MP(out[v],-1)) -H[d].begin() -1; //cout<<l<<" "<<r<<" gg "<<H[d][l].FT<<" "<<H[d][r].FT<<endl; int t=H[d][l].SD ^ H[d][r].SD; if(!(t-(t & -t))) puts("Yes"); else puts("No"); } //system("pause"); return 0;}
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