后续:

点分治标程

使用father数组

比使用vis数组优秀(不需要对vis初始化)

https://codeforces.com/problemset/problem/1174/F

https://codeforces.com/blog/entry/67388

有助于理解树链剖分 和 点分治

题解写得挺好

重链

重链中的点的子树的大小是最大的

重链外的点作为根节点的 子树 大小 < 1/2总的点数目

每次处理后到达重链外的点(若是重链内的点，判断结束)

 #include <cstdio>

 #include <cstdlib>

 #include <cmath>

 #include <cstring>

 #include <string>

 #include <algorithm>

 #include <iostream>

 using namespace std;

 #define ll long long

 const double eps=1e-;

 const ll inf=1e9;

 const ll mod=1e9+;

 const int maxn=2e5+;

 struct node

 {

     int d;

     node *to;

 }*e[maxn];

 char str[];

 int siz[maxn],fa[maxn],link[maxn],cnt_link;

 void dfs(int d)

 {

     node *p=e[d];

     siz[d]=;

     while (p)

     {

         if (fa[d]!=p->d)

         {

             fa[p->d]=d;

             dfs(p->d);

             siz[d]+=siz[p->d];

         }

         p=p->to;

     }

 }

 void findleaf(int d)

 {

     node *p;

     int num;

     cnt_link=;

     link[]=d;

     while ()

     {

         p=e[d];

         num=;

         while (p)

         {

             if (fa[d]!=p->d)

             {

                 if (siz[p->d]>siz[num])

                     num=p->d;

             }

             p=p->to;

         }

         if (!num)

             break;

         link[++cnt_link]=num;

         d=num;

     }

 }

 int main()

 {

     node *p;

     int n,x,y,i,root,dist,r,pos;

     scanf("%d",&n);

     for (i=;i<n;i++)

     {

         scanf("%d%d",&x,&y);

         p=new node();

         p->d=y;

         p->to=e[x];

         e[x]=p;

         p=new node();

         p->d=x;

         p->to=e[y];

         e[y]=p;

     }

     gets(str);

     printf("d %d\n",);

     fflush(stdout);

     scanf("%d",&dist);

     if (dist==)

     {

         printf("! %d\n",);

         fflush(stdout);

         return ;

     }

     root=;

     while ()

     {

         dfs(root);

         findleaf(root);

         printf("d %d\n",link[cnt_link]);

         fflush(stdout);

         scanf("%d",&r);

         pos=(dist+cnt_link-r)/;

         root=link[pos];

         dist=dist-pos;

         if (dist==)

         {

             printf("! %d\n",root);

             fflush(stdout);

             return ;

         }

         printf("s %d\n",root);

         fflush(stdout);

         scanf("%d",&root);

         dist--;

         if (dist==)

         {

             printf("! %d\n",root);

             fflush(stdout);

             return ;

         }

     }

     return ;

 }

 /*

 6

 1 2

 2 3

 3 4

 4 5

 5 6

 7

 1 2

 1 3

 2 4

 2 5

 3 6

 3 7

 7

 1 2

 1 3

 2 4

 2 5

 3 6

 3 7

 d 1

 2

 d 7

 4

 s 1

 2

 d 7

 */

树重心(点分治)

作为树重心的点，若干个子树

max(size of subtree) 最小

子树的大小均小于 < 1/2总的点数目

每次处理后到达新当前树重心的子树(若是树重心，判断结束)

 #include <cstdio>

 #include <cstdlib>

 #include <cmath>

 #include <cstring>

 #include <string>

 #include <algorithm>

 #include <iostream>

 using namespace std;

 #define ll long long

 const double eps=1e-;

 const ll inf=1e9;

 const ll mod=1e9+;

 const int maxn=2e5+;

 /*

 fa

 用于cdq分治比较好

 */

 struct node

 {

     int d;

     node *to;

 }*e[maxn];

 char str[];

 int siz[maxn],fa[maxn],dep[maxn],minsiz,root,pre_root,nn;

 bool hav[maxn];

 void dfs(int d)

 {

     int maxs=;

     node *p=e[d];

     siz[d]=;

     while (p)

     {

         if (fa[d]!=p->d && !hav[p->d])

         {

             fa[p->d]=d;

             dep[p->d]=dep[d]+;

             dfs(p->d);

             siz[d]+=siz[p->d];

             maxs=max(maxs,siz[p->d]);

         }

         p=p->to;

     }

     maxs=max(maxs,nn-siz[d]);

     if (maxs<minsiz)

         minsiz=maxs,root=d;

 }

 int main()

 {

     node *p;

     int n,x,y,i,r,dist;

     scanf("%d",&n);

     for (i=;i<n;i++)

     {

         scanf("%d%d",&x,&y);

         p=new node();

         p->d=y;

         p->to=e[x];

         e[x]=p;

         p=new node();

         p->d=x;

         p->to=e[y];

         e[y]=p;

     }

     gets(str);

     printf("d %d\n",);

     fflush(stdout);

     scanf("%d",&dist);

     if (dist==)

     {

         printf("! %d\n",);

         fflush(stdout);

         return ;

     }

     nn=n;

     root=;

     while ()

     {

         pre_root=root;

         dep[pre_root]=;

         minsiz=inf;

         dfs(root);

         if (root==pre_root)

         {

             printf("s %d\n",root);

             fflush(stdout);

             hav[root]=;

             scanf("%d",&root);

             dist--;

             if (dist==)

             {

                 printf("! %d\n",root);

                 fflush(stdout);

                 return ;

             }

             nn=siz[root];

         }

         else

         {

             printf("d %d\n",root);

             fflush(stdout);

             scanf("%d",&r);

             if (r==)

             {

                 printf("! %d\n",root);

                 fflush(stdout);

                 return ;

             }

             hav[root]=;

             if (r+dep[root]==dist)

             {

                 printf("s %d\n",root);

                 fflush(stdout);

                 dist-=dep[root]+;

                 scanf("%d",&root);

                 if (dist==)

                 {

                     printf("! %d\n",root);

                     fflush(stdout);

                     return ;

                 }

                 nn=siz[root];

             }

             else

             {

                 nn=siz[pre_root]-siz[root];

                 root=pre_root;///so that no need to initialize array fa

                 ///dist not change

             }

         }

     }

     return ;

 }

 /*

 6

 1 2

 2 3

 3 4

 4 5

 5 6

 7

 1 2

 1 3

 2 4

 2 5

 3 6

 3 7

 15

 1 2

 1 3

 2 4

 2 5

 3 6

 3 7

 4 8

 4 9

 5 10

 5 11

 6 12

 6 13

 7 14

 7 15

 7

 1 2

 2 3

 3 4

 4 5

 3 6

 3 7

 12

 1 2

 2 3

 2 4

 4 5

 4 6

 1 7

 7 8

 8 9

 9 10

 9 11

 11 12

 7

 1 2

 1 3

 1 4

 1 5

 1 6

 6 7

 9

 1 2

 2 3

 3 4

 4 5

 1 6

 1 7

 1 8

 1 9

 8

 1 2

 2 3

 3 4

 2 5

 5 6

 5 7

 5 8

 */

为了省去vis初始化

way1:

use father vex

way2:

vis[d]=1;

...

vis[d]=0;

     #include <cstdio>

     #include <cstdlib>

     #include <cmath>

     #include <cstring>

     #include <string>

     #include <algorithm>

     #include <iostream>

     using namespace std;

     #define ll long long

     const double eps=1e-;

     const ll inf=1e9;

     const ll mod=1e9+;

     const int maxn=2e5+;

     struct node

     {

         int d;

         node *to;

     }*e[maxn];

     char str[];

     int siz[maxn],fa[maxn],link[maxn],cnt_link;

     void dfs(int d)

     {

         node *p=e[d];

         siz[d]=;

         while (p)

         {

             if (fa[d]!=p->d)

             {

                 fa[p->d]=d;

                 dfs(p->d);

                 siz[d]+=siz[p->d];

             }

             p=p->to;

         }

     }

     void findleaf(int d)

     {

         node *p;

         int num;

         cnt_link=;

         link[]=d;

         while ()

         {

             p=e[d];

             num=;

             while (p)

             {

                 if (fa[d]!=p->d)

                 {

                     if (siz[p->d]>siz[num])

                         num=p->d;

                 }

                 p=p->to;

             }

             if (!num)

                 break;

             link[++cnt_link]=num;

             d=num;

         }

     }

     int main()

     {

         node *p;

         int n,x,y,i,root,dist,r,pos;

         scanf("%d",&n);

         for (i=;i<n;i++)

         {

             scanf("%d%d",&x,&y);

             p=new node();

             p->d=y;

             p->to=e[x];

             e[x]=p;

             p=new node();

             p->d=x;

             p->to=e[y];

             e[y]=p;

         }

         gets(str);

         printf("d %d\n",);

         fflush(stdout);

         scanf("%d",&dist);

         if (dist==)

         {

             printf("! %d\n",);

             fflush(stdout);

             return ;

         }

         root=;

         while ()

         {

             fa[root]=;

             dfs(root);

             findleaf(root);

             printf("d %d\n",link[cnt_link]);

             fflush(stdout);

             scanf("%d",&r);

             if (r==)

             {

                 printf("! %d\n",link[cnt_link]);

                 fflush(stdout);

                 return ;

             }

             pos=(dist+cnt_link-r)/;

             root=link[pos];

             dist=dist-pos;

             if (dist==)

             {

                 printf("! %d\n",root);

                 fflush(stdout);

                 return ;

             }

             printf("s %d\n",root);

             fflush(stdout);

             scanf("%d",&root);

             dist--;

             if (dist==)

             {

                 printf("! %d\n",root);

                 fflush(stdout);

                 return ;

             }

         }

         return ;

     }

     /*

     6

     1 2

     2 3

     3 4

     4 5

     5 6

     */
     #include <cstdio>

     #include <cstdlib>

     #include <cmath>

     #include <cstring>

     #include <string>

     #include <algorithm>

     #include <iostream>

     using namespace std;

     #define ll long long

     const double eps=1e-;

     const ll inf=1e9;

     const ll mod=1e9+;

     const int maxn=2e5+;

     struct node

     {

         int d;

         node *to;

     }*e[maxn];

     char str[];

     int siz[maxn],fa[maxn],link[maxn],cnt_link;

     bool vis[maxn];

     void dfs(int d)

     {

         node *p=e[d];

         vis[d]=;

         siz[d]=;

         while (p)

         {

             if (!vis[p->d])

             {

                 dfs(p->d);

                 siz[d]+=siz[p->d];

                 fa[p->d]=d;

             }

             p=p->to;

         }

         vis[d]=;

     }

     void findleaf(int d)

     {

         node *p;

         int num;

         cnt_link=;

         link[]=d;

         while ()

         {

             p=e[d];

             num=;

             while (p)

             {

                 if (fa[d]!=p->d)

                 {

                     if (siz[p->d]>siz[num])

                         num=p->d;

                 }

                 p=p->to;

             }

             if (!num)

                 break;

             link[++cnt_link]=num;

             d=num;

         }

     }

     int main()

     {

         node *p;

         int n,x,y,i,root,dist,r,pos;

         scanf("%d",&n);

         for (i=;i<n;i++)

         {

             scanf("%d%d",&x,&y);

             p=new node();

             p->d=y;

             p->to=e[x];

             e[x]=p;

             p=new node();

             p->d=x;

             p->to=e[y];

             e[y]=p;

         }

         gets(str);

         printf("d %d\n",);

         fflush(stdout);

         scanf("%d",&dist);

         if (dist==)

         {

             printf("! %d\n",);

             fflush(stdout);

             return ;

         }

         root=;

         while ()

         {

     //        memset(vis,0,sizeof(vis));

             fa[root]=;

             dfs(root);

             findleaf(root);

             printf("d %d\n",link[cnt_link]);

             fflush(stdout);

             scanf("%d",&r);

             if (r==)

             {

                 printf("! %d\n",link[cnt_link]);

                 fflush(stdout);

                 return ;

             }

             pos=(dist+cnt_link-r)/;

             root=link[pos];

             dist=dist-pos;

             if (dist==)

             {

                 printf("! %d\n",root);

                 fflush(stdout);

                 return ;

             }

             printf("s %d\n",root);

             fflush(stdout);

             scanf("%d",&root);

             dist--;

             if (dist==)

             {

                 printf("! %d\n",root);

                 fflush(stdout);

                 return ;

             }

         }

         return ;

     }

     /*

     6

     1 2

     2 3

     3 4

     4 5

     5 6

     */

Codeforces Round #563 (Div. 2) F. Ehab and the Big Finale的更多相关文章

  1. Codeforces Round #525 (Div. 2) F. Ehab and a weird weight formula

    F. Ehab and a weird weight formula 题目链接:https://codeforces.com/contest/1088/problem/F 题意: 给出一颗点有权值的树 ...

  2. Codeforces Round #563 (Div. 2) E. Ehab and the Expected GCD Problem

    https://codeforces.com/contest/1174/problem/E dp 好题 *(if 满足条件) 满足条件 *1 不满足条件 *0 ///这代码虽然写着方便,但是常数有点大 ...

  3. Codeforces Round #563 (Div. 2) C. Ehab and a Special Coloring Problem

    链接:https://codeforces.com/contest/1174/problem/C 题意: You're given an integer nn. For every integer i ...

  4. Codeforces Round #563 (Div. 2) B. Ehab Is an Odd Person

    链接:https://codeforces.com/contest/1174/problem/B 题意: You're given an array aa of length nn. You can ...

  5. Codeforces Round #563 (Div. 2) A. Ehab Fails to Be Thanos

    链接:https://codeforces.com/contest/1174/problem/A 题意: You're given an array aa of length 2n2n. Is it ...

  6. Codeforces Round #563 (Div. 2)/CF1174

    Codeforces Round #563 (Div. 2)/CF1174 CF1174A Ehab Fails to Be Thanos 其实就是要\(\sum\limits_{i=1}^n a_i ...

  7. Codeforces Round #485 (Div. 2) F. AND Graph

    Codeforces Round #485 (Div. 2) F. AND Graph 题目连接: http://codeforces.com/contest/987/problem/F Descri ...

  8. Codeforces Round #486 (Div. 3) F. Rain and Umbrellas

    Codeforces Round #486 (Div. 3) F. Rain and Umbrellas 题目连接: http://codeforces.com/group/T0ITBvoeEx/co ...

  9. Codeforces Round #501 (Div. 3) F. Bracket Substring

    题目链接 Codeforces Round #501 (Div. 3) F. Bracket Substring 题解 官方题解 http://codeforces.com/blog/entry/60 ...

随机推荐

  1. js中常用的正则表达式

    我一般对正则的使用方式如下,该方法会返回一个boolean值,然后对这个返回值来进行判断 // 判断是否是整数 function isInt(num) { var reg = new RegExp(& ...

  2. rmq +二分暴力 hdu 5726

    参考博客 题意:n 个数字的数列,有m个询问:求出  L   到   R 的  gcd(最大公约数 ),然后问这整个序列中有多少个区间的  gcd  和这个一样. 分析:L 到  R的gcd直接用RM ...

  3. 多flavor导致的源码依赖问题-- Cannot choose between the following configurations of project :sample:

    一.背景: 当我们在源码依赖的时候经常会导致一些问题. 我们的主工程有如下配置,它依赖了一个sample的本地工程 flavorDimensions "demo" productF ...

  4. 转载：vs2010 问题 >LINK : fatal error LNK1123: 转换到 COFF 期间失败: 文件无效或损坏

    原文链接:http://www.cnblogs.com/newpanderking/articles/3372969.html >LINK : fatal error LNK1123: 转换到 ...

  5. apache + php 无法访问redis

    1.在有扩展的情况下 2.测试连接 <?php $redis=new Redis(); $redis->connect('127.0.0.1',6379); echo "succ ...

  6. linux fcntl 对文件描述符控制

    linux fcntl 对文件描述符控制 linux fcntl 对文件描述符控制 linux fcntl 对文件描述符控制

  7. boost 实现读写锁

    #include <boost/thread/shared_mutex.hpp> #include <boost/thread/locks.hpp> using BoostMu ...

  8. 安装memcached报错：If it's already installed, specify its path using --with-libevent=/dir/

    一.安装memcached,执行./configure --prefix=/usr/local/memcached时候报错: 问题:If it's already installed, specify ...

  9. MongoDB启动时遇到的问题

    问题一:child process failed, exited with error number 100 原因分析:100是锁了 解决办法:#cd  /usr/local/mongodb/data ...

  10. 线程池 一 ScheduledThreadPoolExecutor

    java.util.concurrent public class ScheduledThreadPoolExecutor extends ThreadPoolExecutor implements ...