Distance Queries
Time Limit: 2000ms
Memory Limit: 30000KB
This problem will be judged on PKU. Original ID: 198664-bit integer IO format: %lld Java class name: Main
Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in "Navigation Nightmare",followed by a line containing a single integer K, followed by K "distance queries". Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ's distance queries as quickly as possible!
Input
* Lines 1..1+M: Same format as "Navigation Nightmare"
* Line 2+M: A single integer, K. 1 <= K <= 10,000
* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.
* Line 2+M: A single integer, K. 1 <= K <= 10,000
* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.
Output
* Lines 1..K: For each distance query, output on a single line an integer giving the appropriate distance.
Sample Input
7 6 1 6 13 E 6 3 9 E 3 5 7 S 4 1 3 N 2 4 20 W 4 7 2 S 3 1 6 1 4 2 6
Sample Output
13 3 36
Source
解题：LCA求树上任意两点间的距离。任意两点间只有一条路啊！不然那还是树么？^_^。。。。
LCA(a,b) = c所以d(a,b) = d(a,root)+d(b,root)-2*d(c,root);人字形？嘻嘻！傻逼。。。。。当时我居然不能明白这个
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long 10 #define INF 0x3f3f3f 11 using namespace std; 12 const int maxn = 100010; 13 struct arc{ 14 int to,w; 15 }; 16 struct query{ 17 int to,id; 18 }; 19 int n,m,k,ans[maxn],d[maxn],uf[maxn]; 20 bool vis[maxn]; 21 vector<arc>g[maxn]; 22 vector<query>q[maxn]; 23 int Find(int x){ 24 if(x != uf[x]) 25 uf[x] = Find(uf[x]); 26 return uf[x]; 27 } 28 void tarjan(int u,int ds){ 29 vis[u] = true; 30 d[u] = ds; 31 uf[u] = u; 32 int i; 33 for(i = 0; i < g[u].size(); i++){ 34 if(!vis[g[u][i].to]) {tarjan(g[u][i].to,ds+g[u][i].w);uf[g[u][i].to] = u;} 35 } 36 for(i = 0; i < q[u].size(); i++) 37 if(vis[q[u][i].to]){ 38 ans[q[u][i].id] = d[u]+d[q[u][i].to]-2*d[Find(q[u][i].to)]; 39 } 40 } 41 int main(){ 42 int i,j,u,v,w; 43 char ch; 44 while(~scanf("%d %d",&n,&m)){ 45 for(i = 0; i <= n; i++){ 46 g[i].clear(); 47 q[i].clear(); 48 d[i] = 0; 49 vis[i] = false; 50 } 51 for(i = 0; i < m; i++){ 52 scanf("%d %d %d %c",&u,&v,&w,&ch); 53 g[u].push_back((arc){v,w}); 54 g[v].push_back((arc){u,w}); 55 } 56 scanf("%d",&k); 57 for(i = 0; i < k; i++){ 58 scanf("%d %d",&u,&v); 59 q[u].push_back((query){v,i}); 60 q[v].push_back((query){u,i}); 61 } 62 tarjan(1,0); 63 for(i = 0; i < k; i++) 64 printf("%d ",ans[i]); 65 } 66 return 0; 67 }