Codeforces 813C The Tag Game
来源:互联网 发布:linux启动图形化命令 编辑:程序博客网 时间:2024/05/21 10:14
理解题
// 38#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#include <list>#include <deque>#include <map>#include <queue>using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int INF = 1000000000;const int maxn = 200010;int T,n,m;vector <int> g[maxn];int level[maxn];int fa[maxn];int res=0;void dfs (int s,int f,int t){ int i; level[s] = t; fa[s]= f; for (i=0;i<g[s].size ();i++){ int v= g[s][i]; if (v==f) continue; dfs (v,s,t+1); } return;}void dfs2 (int s,int t){ int i; res = max (res,t); for (i=0;i<g[s].size ();i++){ int v = g[s][i]; if (level[v]<level[s]) continue; dfs2 (v,t+1); } return;}int main (){ int n,x; int i,j; scanf ("%d %d",&n,&x); for (i=1;i<n;i++){ int a,b; scanf ("%d %d",&a,&b); g[a].push_back (b); g[b].push_back (a); } dfs (1,0,1); int t; if ((1+x)%2==0){ t = (1+x)/2-1; } else t = (1+x)/2; int v= x; for (i=2;i<=t;i++){ v =fa[x]; } //printf("%d\n",v); res = 0; dfs2 (v,0); //printf("%d\n",res); printf("%d\n", 2*(level[v]-level[1]+res) ); return 0;}
阅读全文
0 0
- codeforces 813C The Tag Game
- codeforces 813C. The Tag Game
- Codeforces 813 C The Tag Game
- Codeforces 813C The Tag Game 题解
- codeforces 813C The Tag Game dfs
- codeforces 813C The Tag Game
- codeforces 813C The Tag Game
- Codeforces 813C The Tag Game
- Codeforces 813C The Tag Game【思维+Dfs】
- codeforces 813C The Tag Game dfs && 路径长度
- The Tag Game CodeForces
- The Tag Game(CodeForces
- C. The Tag Game
- Educational Codeforces Round 22 C. The Tag Game dfs
- Educational Codeforces Round 22-C. The Tag Game-搜索,贪心
- Educational Codeforces Round 22 C. The Tag Game(思维)
- Educational Codeforces Round 22 C. The Tag Game 搜索
- Codeforces Round 22 C. The Tag Game ( 搜索
- 阿里的盔甲、未来20年发展的动力以及对未来的洞察
- LintCode-分治-合并k个排序链表
- 想学区块链技术?来这!
- 51Nod 拉勾第一题(数位dp)
- Mesos容器引擎的架构设计和实现解析
- Codeforces 813C The Tag Game
- AI 线上峰会 | 人工智能技术解析与实战
- LIS算法
- java基础(一)
- centos rsyslog配置及简单应用
- 万字长文|深度剖析Service Mesh服务网格新生代Istio
- 国内外主流移动支付方案与TEE
- 避免大规模故障的微服务架构设计之道
- 网易数据运河系统NDC设计与应用