hdoj 2586 How far away ？ 【LCA转RMQ入门题】
How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8693 Accepted Submission(s): 3044
Problem Description
There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.
Input
First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
Output
For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
Sample Input
2 3 2 1 2 10 3 1 15 1 2 2 3 2 2 1 2 100 1 2 2 1
Sample Output
10 25 100 100
题意：给你一个N个点的树和Q次查询，每次查询问你任意两点间距离。
LCA转RMQ 点这里：
点我
思路：<u, v>距离 = dist[ u ] + dist[ v ] - 2 * dist[ LCA(u，v)]。其中dist存储节点到根的距离。
AC代码：注意手动开栈 。。。
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 40000+100
#define MAXM 100000+1000
#define LL long long
using namespace std;
struct Edge
{
int from, to, val, next;
};
Edge edge[MAXM];
int head[MAXN], edgenum;
int vs[MAXN<<1];
int depth[MAXN<<1];
int id[MAXN];
int dfs_clock;
int dist[MAXN];
int N, Q;
void init()
{
edgenum = 0;
memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int w)
{
Edge E = {u, v, w, head[u]};
edge[edgenum] = E;
head[u] = edgenum++;
}
void getMap()
{
int a, b, c;
for(int i = 1; i < N; i++)
{
scanf("%d%d%d", &a, &b, &c);
addEdge(a, b, c), addEdge(b, a, c);
}
}
void DFS(int u, int fa, int d)
{
id[u] = dfs_clock;
vs[dfs_clock] = u;
depth[dfs_clock++] = d;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(v == fa) continue;
dist[v] = dist[u] + edge[i].val;
DFS(v, u, d+1);
vs[dfs_clock] = u;
depth[dfs_clock++] = d;
}
}
void find_depth()
{
memset(id, 0, sizeof(id));
memset(vs, 0, sizeof(vs));
memset(depth, 0, sizeof(depth));
memset(dist, 0, sizeof(dist));
dfs_clock = 1;
DFS(1, -1, 0);
}
int dp[MAXN<<1][30];
void RMQ_init(int NN)
{
for(int i = 1; i <= NN; i++)
dp[i][0] = i;
for(int j = 1; (1<<j) <= NN; j++)
{
for(int i = 1; i + (1<<j) - 1 <= NN; i++)
{
int a = dp[i][j-1];
int b = dp[i + (1<<(j-1))][j-1];
if(depth[a] <= depth[b])
dp[i][j] = a;
else
dp[i][j] = b;
}
}
}
int query(int L, int R)
{
int k = 0;
while((1<<(k+1)) <= R-L+1) k++;
int a = dp[L][k];
int b = dp[R - (1<<k) + 1][k];
if(depth[a] <= depth[b])
return a;
else
return b;
}
int LCA(int u, int v)
{
int x = id[u];
int y = id[v];
if(x < y)
return vs[query(x, y)];
else
return vs[query(y, x)];
}
void solve()
{
int a, b;
while(Q--)
{
scanf("%d%d", &a, &b);
printf("%d\n", dist[a] + dist[b] - 2 * dist[LCA(a, b)]);
}
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &N, &Q);
init();
getMap();
find_depth();
RMQ_init(dfs_clock - 1);
solve();
}
return 0;
}