链接：http://wikioi.com/problem/1036/
题意不写了。
思路：很明显找到lca然后用两个点的深度相加-lca的深度就是这一步的最近步数。
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <algorithm> 5 #include <stdlib.h> 6 #include <vector> 7 #include <queue> 8 #define loop(s,i,n) for(i = s;i < n;i++) 9 #define cl(a,b) memset(a,b,sizeof(a)) 10 using namespace std; 11 const int maxn = 100005; 12 int low[maxn],dfn[maxn],set[maxn],father[maxn],dfsclock,cut; 13 int depth[maxn]; 14 15 vector<int >g[maxn]; 16 int find(int x) 17 { 18 if(set[x] != x) 19 set[x] = find(set[x]); 20 21 return set[x]; 22 } 23 int merge(int x,int y) 24 { 25 x = find(x); 26 y = find(y); 27 if(y != x) 28 { 29 set[y] = x; 30 return 1; 31 } 32 return 0; 33 } 34 void tarjan(int u,int pre,int deep) 35 { 36 int v,i,j; 37 dfn[u] = low[u] = ++dfsclock; 38 depth[u] = deep; 39 loop(0,i,g[u].size()) 40 { 41 v = g[u][i]; 42 if(!dfn[v]) 43 { 44 tarjan(v,u,deep+1); 45 father[v] = u; 46 low[u] = min(low[v],low[u]); 47 if(low[v] > dfn[u]) 48 cut++; 49 else 50 merge(u,v); 51 } 52 else if(v != pre) 53 low[u] = min(low[u],dfn[v]); 54 } 55 } 56 int lca(int u,int v) 57 { 58 59 while(u != v) 60 { 61 while(dfn[u] >= dfn[v] && u != v) 62 { 63 if(merge(u,father[u])) 64 cut--; 65 u = father[u]; 66 } 67 while(dfn[v] >= dfn[u] && u != v) 68 { 69 if(merge(v,father[v])) 70 cut--; 71 v = father[v]; 72 } 73 } 74 return u; 75 } 76 int main() 77 { 78 int n,m; 79 int i,x,y; 80 scanf("%d",&n); 81 { 82 int u,v; 83 loop(0,i,n+1) 84 { 85 g[i].clear(); 86 set[i] = i; 87 father[i] = 0; 88 } 89 loop(1,i,n) 90 { 91 scanf("%d %d",&u,&v); 92 g[u].push_back(v); 93 g[v].push_back(u); 94 95 } 96 cl(dfn,0); 97 cl(low,0); 98 cut = dfsclock = 0; 99 100 int k; 101 scanf("%d",&k); 102 tarjan(1,-1,0); 103 u = 1; 104 int ans; 105 ans = 0; 106 while(k--) 107 { 108 scanf("%d",&v); 109 int f; 110 f = lca(u,v); 111 ans += depth[u]+depth[v]-2*depth[f]; 112 u = v; 113 } 114 cout<<ans<<endl; 115 } 116 return 0; 117 }