Codeforces Round #316 (Div. 2) D. Tree Requests (DFS序)
来源:互联网 发布:远程监控域名 编辑:程序博客网 时间:2024/05/29 11:10
题目地址:http://codeforces.com/contest/570/problem/D
比赛的时候实在没想到DFS序,。。想到DFS序后,分别存起每个深度的所有节点的DFS序,处理出前缀异或和,然后二分找到两个端点,再异或一下,就求出了所求区间的异或和,由于偶数次的都被异或掉了,所以判断下奇数次数是否大于1即可。
代码如下:
#include <iostream>#include <string.h>#include <math.h>#include <queue>#include <algorithm>#include <stdlib.h>#include <map>#include <set>#include <stdio.h>#include <time.h>using namespace std;#define LL long long#define pi acos(-1.0)#pragma comment(linker, "/STACK:1024000000")const LL mod=3221225473;const int INF=0x3f3f3f3f;const double eqs=1e-9;const int MAXN=500000+10;vector<int>vec[MAXN];vector<int>::iterator it;int in[MAXN], out[MAXN], sum[MAXN];int head[MAXN], cnt, now;char s[MAXN];struct node{ int u, v, next;}edge[MAXN];void add(int u, int v){ edge[cnt].v=v; edge[cnt].next=head[u]; head[u]=cnt++;}void init(int n){ memset(head,-1,sizeof(head)); cnt=now=0; memset(sum,0,sizeof(sum)); for(int i=1;i<=n;i++){ vec[i].clear(); vec[i].push_back(0); }}void dfs(int u, int h){ now++; sum[now]=sum[*(--vec[h].end())]^(1<<s[u]-'a'); vec[h].push_back(now); in[u]=now; for(int i=head[u];i!=-1;i=edge[i].next){ int v=edge[i].v; dfs(v,h+1); } out[u]=now;}int main(){ int n, m, i, j, u, h, ans, l, r, flag; while(scanf("%d%d",&n,&m)!=EOF){ init(n); for(i=2;i<=n;i++){ scanf("%d",&u); add(u,i); } scanf("%s",s+1); dfs(1,1); while(m--){ scanf("%d%d",&u,&h); it=lower_bound(vec[h].begin(),vec[h].end(),out[u]); if((*it)!=out[u]) it--; r=sum[*it]; it=--lower_bound(vec[h].begin(),vec[h].end(),in[u]); l=sum[*it]; ans=r^l; flag=0; for(i=0;i<26;i++){ if(ans&(1<<i)) flag++; if(flag==2) break; } if(flag==2) puts("No"); else puts("Yes"); } } return 0;}
2 0
- Codeforces Round #316 (Div. 2) D. Tree Requests (DFS序)
- Codeforces Round #316 (Div. 2) D. Tree Requests(DFS序+BFS+二分)
- Codeforces Round #316 (Div. 2) D. Tree Requests(DFS+状态压缩)
- Codeforces Round #316 (Div. 2)-D. Tree Requests-DFS+二分+hash
- *Codeforces Round #316 (Div. 2)- D. Tree Requests (dfs+二分)
- Codeforces Round #316 (Div. 2) D. Tree Requests
- Codeforces Round #316 (Div. 2) D. Tree Requests dfs_clock,二分
- Codeforces Round #316 (Div. 2) D Tree Requests
- Codeforces Round #316 (Div. 2) D. Tree Requests
- Codeforces Round #316 (Div. 2)D. Tree Requests
- Codeforces Round #316 (Div. 2) D. Tree Requests
- 善用stl。。。。。。。。Codeforces Round #316 (Div. 2) D - Tree Requests
- Codeforces Round #316 (Div. 2) D. Tree Requests
- Tree Requests-Codeforces Round #316 (Div. 2)
- Codeforces Round #316 (Div.2)Tree Requests
- Codeforces Round #316 (Div. 2) D. Tree Requests 树 离线在线 算法
- Codeforces Round #316 D. Tree Requests 树剥分
- codeforces #316 D.Tree Requests (巧妙的dfs序)
- 【小熊刷题】Add Two Numbers in Linked Lists
- [Java]Contains Duplicate II 包含重复数字
- ViewPager+Fragment+RadioGruop实现滑动效果
- 在被管理中体会管理的难!
- SingleTon
- Codeforces Round #316 (Div. 2) D. Tree Requests (DFS序)
- Leetcode#53||Maximum Subarray
- 【HDU4405】【Aeroplane chess】【概率dp】
- Ubiquitous Religions
- Jedis主从切换实现
- 1. JavaScript Array 对象
- HDU1023 Train Problem II【Catalan数】
- CLRS第五章思考题
- table表格td撑开