codeforces 813C The Tag Game

最新推荐文章于 2025-01-15 22:19:39 发布

至此曾经

最新推荐文章于 2025-01-15 22:19:39 发布

阅读量1.7k

点赞数

CC 4.0 BY-SA版权

分类专栏： ###### 搜索 ###### DFS

本文链接：https://blog.youkuaiyun.com/qq_32944513/article/details/72877155

###### 搜索 ###### 同时被 3 个专栏收录

8 篇文章

订阅专栏

DFS

6 篇文章

订阅专栏

codeforce

5 篇文章

订阅专栏

本文探讨了一种名为“标签游戏”的策略游戏，在该游戏中，玩家A从根节点出发，玩家B从非根节点开始，目标是A尽可能快地追上B。文章详细解析了最优策略，包括如何通过寻找特定路径上的最深节点来最大化移动步数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

C. The Tag Game
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Alice got tired of playing the tag game by the usual rules so she offered Bob a little modification to it. Now the game should be played on an undirected rooted tree of n vertices. Vertex 1 is the root of the tree.

Alice starts at vertex 1 and Bob starts at vertex x (x ≠ 1). The moves are made in turns, Bob goes first. In one move one can either stay at the current vertex or travel to the neighbouring one.

The game ends when Alice goes to the same vertex where Bob is standing. Alice wants to minimize the total number of moves and Bob wants to maximize it.

You should write a program which will determine how many moves will the game last.

Input
The first line contains two integer numbers n and x (2 ≤ n ≤ 2·105, 2 ≤ x ≤ n).

Each of the next n - 1 lines contains two integer numbers a and b (1 ≤ a, b ≤ n) — edges of the tree. It is guaranteed that the edges form a valid tree.

Output
Print the total number of moves Alice and Bob will make.

Examples
input
4 3
1 2
2 3
2 4
output
4
input
5 2
1 2
2 3
3 4
2 5
output

题意：一颗无向有根树，A在根节点1，B在非根节点x。B先走然后再A走(如此下去)，每一步要么向相连点移动要么不动。A要以尽可能减少移动步数的走法，B以尽可能增加步数的走法。当A与B相遇时游戏结束。求最大的移动步数。

分析：B肯定在A的子树里所以A走法就是不走多余的节点直接去接近B，而B可以找到他能走到的离根节点1距离最大的点这样就能使最后步数最大。

所以求出刚开始A·B的路径上的节点，找到此路径上深度最大的节点且B能走到的节点Y，ans =( 1到Y+Y的能到的最深深度)*2.

暂时没有想到更简单的想法和代码的写法，直接贴上代码：

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#define siz 201015
using namespace std;
int n,x,cot;
int d[siz],path[siz],vis[siz],dep[siz];
vector<int>G[siz];
int dfs(int u,int fa)//求深度
{
    int len = G[u].size(),Max = 0;
    for(int i=0; i<len; i++)
    {
        int v = G[u][i];
        if(v == fa) continue;
        dep[v] = dep[u] + 1;
        Max = max(dfs(v,u),Max);
    }
    return d[u] = Max+1;
}
void pdfs(int u,int fa,int t)//求出路径
{
    if(cot) return;
    path[t] = u;
    if(u == x)
    {
        cot = t;
        return;
    }
    int len = G[u].size();
    for(int i=0; i<len; i++)
    {
        int v = G[u][i];
        if(v == fa) continue;
        pdfs(v,u,t+1);
    }
}
void solve()
{
    int ans = 0;
    memset(d,0,sizeof(d));
    memset(vis,0,sizeof(vis));
    cot = 0,dep[1] = 0;
    dfs(1,-1);
    pdfs(1,-1,1);
    for(int i=cot,j=1; i; i--,j++)//B先走 然后再A走，vis[i] 表示A所在的位置。
    {
        vis[j] = 1;
        if(!vis[i]) ans = max(ans,d[path[i]]+dep[path[i]]-1);
    }
    cout<<ans*2<<endl;
}
int main()
{
    int u,v;
    while(~scanf("%d %d",&n,&x))
    {
        for(int i=1; i<=n; i++) G[i].clear();
        for(int i=1; i<n; i++)
        {
            scanf("%d %d",&u,&v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        solve();
    }
    return 0;
}