POJ1986-Distance Queries（LCA）

Distance Queries

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 13987		Accepted: 4924
Case Time Limit: 1000MS

Description

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. 

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

Hint

Farms 2 and 6 are 20+3+13=36 apart. 

Source

USACO 2004 February

题意：给你一棵树，数的每条边都有边权，有q次询问，查询出两点之间的距离

解题思路：LCA

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <functional>
#include <climits>

using namespace std;

#define LL long long
const int INF = 0x7FFFFFFF;

int f[50009][20],dp[50009][20];
int s[50009], nt[50009 * 2], e[50009 * 2], w[50009 * 2],deep[50009],vis[50009];
char ch[5];

void dfs(int k)
{
vis[k] = 1;
for (int i = 1; i < 19; i++)
{
if (deep[k] < (1 << i)) break;
dp[k][i] = dp[k][i - 1] + dp[f[k][i - 1]][i - 1];
f[k][i] = f[f[k][i - 1]][i - 1];
}
for (int i = s[k]; ~i; i = nt[i])
{
int ee = e[i];
if (vis[ee]) continue;
deep[ee] = deep[k] + 1,f[ee][0] = k,dp[ee][0] = w[i];
dfs(ee);
}
}

int lca(int x, int y)
{
if (deep[x] < deep[y]) swap(x, y);
int k = deep[x] - deep[y],ans=0;
for (int i = 0; i < 20; i++)
if (k&(1 << i)) ans += dp[x][i], x = f[x][i];
for (int i = 19; i >= 0; i--)
if (f[x][i] != f[y][i]) ans = ans + dp[x][i] + dp[y][i], x = f[x][i], y = f[y][i];
if (x == y) return ans;
return ans + dp[x][0] + dp[y][0];
}

int main()
{
int n, q;
while (~scanf("%d%d", &n, &q))
{
int u, v, ww,cnt=1;
memset(s, -1, sizeof s);
for (int i = 1; i < n; i++)
{
scanf("%d%d%d%s", &u, &v, &ww,ch);
nt[cnt] = s[u], s[u] = cnt, e[cnt] = v, w[cnt++] = ww;
nt[cnt] = s[v], s[v] = cnt, e[cnt] = u, w[cnt++] = ww;
}
deep[1] = 0;
memset(vis, 0, sizeof vis);
memset(f, 0, sizeof f);
memset(dp, 0, sizeof dp);
dfs(1);
scanf("%d", &q);
while (q--)
{
int x, y;
scanf("%d%d", &x, &y);
printf("%d\n",lca(x, y));
}
}
return 0;
}