poj 1986 Distance Queries (lca,在线做法dfs+rmq)
题目链接 题意:求树上两点的最短距离? 思路: dis[i]表示点i到根节点的距离,那么任意两点u,v的最短距离d = dis[u]+dis[v]-2*dis[LCA(u,v)]. 只需要求出rmq+dfs的在线方法求出lca(u,v)即可。
/* ***********************************************
Author :111qqz
Created Time :2016年05月20日 星期五 15时36分47秒
File Name :code/poj/1986.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=4E4+7;
7int n,m;
8vector < pi > edge[N];
9int q;
10int in[N];
11int E[2*N],R[2*N],dis[N],depth[2*N];
12int p;
13int dp[2*N][20];
14void dfs( int u,int dep,int d,int pre)
15{
1 // cout<<"u:"<<u<<" dep:"<<dep<<" d:"<<d<<endl;
2 p++;
3 E[p] = u;
4 depth[p] = dep;
5 R[u] = p ;
6 dis[u] = d;
1 int siz = edge[u].size();
2 for ( int i = 0 ; i < siz ; i++)
3 {
4 int v = edge[u][i].fst;
5 if (v==pre) continue;
6 dfs(v,dep+1,d+edge[u][i].sec,u);
1 p++;
2 E[p] = u;
3 depth[p] = dep;
4 }
5}
1int _min( int l,int r)
2{
3 if (depth[l]<depth[r]) return l;
4 return r;
5}
6void rmq_init()
7{
8 for ( int i = 1 ; i <= 2*n+2 ; i++) dp[i][0] = i;
1 for ( int j = 1 ; (1<<j) <= 2*n+2 ; j++)
2 for ( int i = 1 ; i + (1<<j)-1 <= 2*n+2 ; i++)
3 dp[i][j] = _min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
4}
1int rmq_min( int l,int r)
2{
3if (l>r) swap(l,r);
4int k = 0 ;
5while (1<<(k+1)<=r-l+1) k++;
6return _min(dp[l][k],dp[r-(1<<k)+1][k]);
7}
8int main()
9{
10 #ifndef ONLINE_JUDGE
11 freopen("code/in.txt","r",stdin);
12 #endif
1 ms(in,0);
2 scanf("%d %d",&n,&m);
3 for ( int i = 1 ; i <= n ; i++) edge[i].clear();
4 for ( int i = 1 ; i <= m ; i++)
5 {
6 int u,v,w;
7 char dir[5];
8 scanf("%d%d%d%s",&u,&v,&w,dir);
9 edge[u].push_back(make_pair(v,w));
10 edge[v].push_back(make_pair(u,w));
11 }
1 p = 0 ;
2 dfs(1,0,0,-1);
3 rmq_init();
1 scanf("%d",&q);
2 while (q--)
3 {
4 int u,v;
5 scanf("%d%d",&u,&v);
6 // cout<<"u:"<<u<<" v:"<<v<<endl;
7 int LCA = E[rmq_min(R[u],R[v])];
8 // cout<<"LCA:"<<LCA<<endl;
9 int ans;
10 ans = dis[u]+dis[v]-2*dis[LCA];
11 printf("%d\n",ans);
12 }
13 #ifndef ONLINE_JUDGE
14 fclose(stdin);
15 #endif
16 return 0;
17}