牛客多校9-K.The Flee Plan of Groundhog

题意：

　　a的速度是一秒走一格，b的速度是一秒走两格，前t秒a沿着a->b的最短路去找b，t秒的时候b反过来抓a，问a最长多少秒才会被抓住

 

做法：

　　先dfs处理出b从n走到图上任意一点所需要的时间。再在t时a所处的位置bfs处理出a从t时位置到图上各点的时间，如果a和b同时到某点或比b晚到某点，则会被b抓住，维护一个被抓时间的最大值即可

 

CODE

 

  1 #include <bits/stdc++.h>
  2 #define dbug(x) cout << #x << "=" << x << endl
  3 #define eps 1e-8
  4 #define pi acos(-1.0)
  5  
  6 using namespace std;
  7 typedef long long LL;
  8  
  9 const int inf = 0x3f3f3f3f;
 10  
 11 template<class T>inline void read(T &res)
 12 {
 13    char c;T flag=1;
 14    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
 15    while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
 16 }
 17 
 18 const int maxn = 1e5 + 7;
 19 
 20 int n, t;
 21 
 22 vector<int> g[maxn];
 23 int fa[maxn];
 24 int depth[maxn], vis_time[maxn];
 25 
 26 int ans;
 27 
 28 void dfs(int x, int pre){ 
 29     int d = g[x].size();
 30     for ( int i = 0; i <= d - 1; ++i ) {
 31         int v = g[x][i];
 32         if(v == pre) {
 33             continue;
 34         }
 35         fa[v] = x;
 36         depth[v] = depth[x] + 1;
 37         vis_time[v] = depth[v] / 2 + depth[v] % 2;
 38         dfs(v, x);
 39     }
 40 }
 41 
 42 bool vis[maxn];
 43 int vis_time2[maxn];
 44 
 45 void bfs(int x) {
 46     queue<int> q;
 47     q.push(x);
 48     while(!q.empty()) {
 49         int temp = q.front();
 50         q.pop();
 51         if(vis[temp])
 52             continue;
 53         vis[temp] = 1;
 54         // dbug(temp);
 55         // printf("::vis[%d]:%d vis2[%d]:%d\n",temp, vis_time[temp], temp, vis_time2[temp]);
 56         if(vis_time2[temp] <= vis_time[temp]) {
 57             ans = max(ans, vis_time[temp]);
 58         }
 59         // dbug(ans);
 60         if(vis_time2[temp] >= vis_time[temp]) {
 61             continue;
 62         }
 63         int d = g[temp].size();
 64         // dbug(d);
 65         for ( int i = 0; i < d; ++i ) {
 66             int v = g[temp][i];
 67             // dbug(v);
 68             if(vis[v]) {
 69                 continue;
 70             }
 71             vis_time2[v] = vis_time2[temp] + 1;
 72             q.push(v);
 73         }
 74     }
 75 }
 76 
 77 int main()
 78 {
 79     read(n); read(t);
 80     for ( int i = 1; i <= n - 1; ++i ) {
 81         int x, y;
 82         read(x); read(y);
 83         g[x].push_back(y);
 84         g[y].push_back(x);
 85     }
 86     dfs(n, n);
 87     int node = 1;
 88     while(t--) {
 89         node = fa[node];
 90     }
 91     bfs(node);
 92     // for ( int i = 1; i <= n; ++i ) {
 93     //     printf("time[%d]:%d time2[%d]:%d\n",i, vis_time[i], i, vis_time2[i]);
 94     // }
 95     cout << ans << endl;
 96     return 0;
 97 }
 98 /*
 99 7
100 1
101 1 2
102 2 4
103 4 5
104 5 6
105 6 3
106 5 7
107 
108 6 1
109 1 2
110 1 3
111 1 4
112 1 5
113 1 6
114 
115 10 2
116 1 2
117 2 5
118 5 7
119 5 6
120 3 6
121 3 4
122 7 8
123 8 9
124 9 10
125 
126 */
View Code

 

posted @ 2020-08-14 15:20  Orangeko  阅读(124)  评论(0)  编辑  收藏  举报