Codeforces Round #805 (Div. 3)G2. Passable Paths

题目大意：

　　给出一个无向无环连通图（树），n个点n-1条边，m次查询，每次询问给出一个集合，问集合里的树是否都在同一条链上（即能否不重复的走一条边而遍历整个点集）

思路：通过求lca，若有三个点x,y,z

　　如果满足dix(x,y)+dix(y,z) == dix(x,z)，说明此时y位于x,z之间，此时他们就在一条链上，只需要枚举除x,y以外剩下的n-2个点

　　判断是否满足类似的条件，每次不断跟新x,y的值使其成为一条链上的两个端点。

  1 # include<iostream>
  2 # include<bits/stdc++.h>
  3 using namespace std;
  4 const int N = 2e5 + 10;
  5 int h[N], e[2 * N], ne[2 * N], idx;
  6 int depth[N], fa[N][25];
  7 int dis[N];
  8 void bfs(int root)/*预处理depth和fa*/
　　 {
  9     memset(depth, 0x3f, sizeof depth);
 10     queue<int> q;
 11     q.push(root);
 12     depth[0] = 0,depth[root] = 1;
 13     while (q.size()) {
 14         int t = q.front();
 15         q.pop();
 16         for (int i = h[t]; i != -1; i = ne[i]) {
 17             int j = e[i];
 18             if (depth[j] > depth[t] + 1) {
 19                 depth[j] = depth[t] + 1;
 20                 dis[j] = dis[t] + 1;
 21                 q.push(j);
 22                 fa[j][0] = t;
 23                 for (int k = 1; k <= 20; ++k) {
 24                     fa[j][k] = fa[fa[j][k - 1]][k - 1];
 25                 }
 26             }
 27         }
 28     }
 29 }
 30 
 31 int lca(int a, int b) {/*倍增求lca（最近公共祖先）*/
 32     if (depth[a] < depth[b]) swap(a, b);
 33     for (int i = 20; i >= 0; --i) {
 34         if (depth[fa[a][i]] >= depth[b]) a = fa[a][i];
 35     }
 36     if (a == b) return a;
 37     for (int i = 20; i >= 0; --i) {
 38         if (fa[a][i] != fa[b][i]) {
 39             a = fa[a][i];
 40             b = fa[b][i];
 41         }
 42     }
 43     return fa[a][0];
 44 }
 45 
 46 void add(int a, int b) {
 47     e[idx] = b;
 48     ne[idx] = h[a];
 49     h[a] = idx++;
 50 }
 51 int getdis(int x, int y) {/*两点之间最短距离*/
 52     return dis[x] + dis[y] - 2 * dis[lca(x, y)];
 53 }
 54 
 55 void solve() {
 56     int n;
 57     bool ok = 1;
 58     cin >> n;
 59     if (n == 1 || n == 2) {
 60         for (int i = 1; i <= n; ++i) {
 61             int x;
 62             cin >> x;
 63         }
 64         cout << "YES" << endl;
 65         return;
 66     }
 67     int  x, y;
 68     cin >> x >> y;
 69     for (int i = 3; i <= n; ++i) {
 70         int z;
 71         cin >> z;
 72         if (getdis(x, z) + getdis(z, y) == getdis(x, y)) continue;
 73         if (getdis(x, y) + getdis(y, z) == getdis(x, z)) {
 74             y = z;
 75             continue;
 76         }
 77         if (getdis(y, x) + getdis(x, z) == getdis(y, z)) {
 78             x = z;
 79             continue;
 80         }
 81         ok = false;
 82     }
 83     if (ok) cout << "YES" << endl;
 84     else cout << "NO" << endl;
 85 }
 86 int tt;
 87 
 88 int main() {
 89 
 90     int n;
 91     cin >> n;
 92     memset(h, -1, sizeof h);
 93     for (int i = 1; i < n; ++i) {
 94         int a, b;
 95         cin >> a >> b;
 96         add(a, b), add(b, a);
 97     }
 98     bfs(1);
 99 //    for(int i = h[1];i!=-1;i = ne[i]) cout<<e[i]<<" ";
100     cin>>tt;
101     while(tt--)solve();
102     return 0;
103 }

标签：Paths,return,G2,getdis,int,Codeforces,fa,depth,dis
来源： https://www.cnblogs.com/empty-y/p/16477281.html