zoj 3195 Design the city (lca,dfs+rmq)

zoj 3195题目链接 题意:求树上三点的最短距离。。。 思路:两两求,和除以2. 因为忘记初始化p=0..WA了将近两个小时。。。? 妈的智障。

  1/* ***********************************************
  2Author :111qqz
  3Created Time :2016年05月21日 星期六 14时44分39秒
  4File Name :code/zoj/3195.cpp
  5************************************************ */
  6
  7
  8
  9#include <cstdio>
 10#include <cstring>
 11#include <iostream>
 12#include <algorithm>
 13#include <vector>
 14#include <queue>
 15#include <set>
 16#include <map>
 17#include <string>
 18#include <cmath>
 19#include <cstdlib>
 20#include <ctime>
 21#define fst first
 22#define sec second
 23#define lson l,m,rt<<1
 24#define rson m+1,r,rt<<1|1
 25#define ms(a,x) memset(a,x,sizeof(a))
 26typedef long long LL;
 27#define pi pair < int ,int >
 28#define MP make_pair
 29
 30using namespace std;
 31const double eps = 1E-8;
 32const int dx4[4]={1,0,0,-1};
 33const int dy4[4]={0,-1,1,0};
 34const int inf = 0x3f3f3f3f;
 35const int N=5E4+7;
 36int n,m;
 37vector < pi > edge[N];
 38int q;
 39int in[N];
 40int E[2*N],R[2*N],dis[N],depth[2*N];
 41int p;
 42int dp[2*N][20];
 43void dfs( int u,int dep,int d,int pre)
 44{
 45
 46  //  cout<<"u:"<<u<<" dep:"<<dep<<" d:"<<d<<endl;
 47    p++;
 48    E[p] = u;
 49    depth[p] = dep;
 50    R[u] = p ;
 51    dis[u] = d;
 52
 53
 54    int siz = edge[u].size();
 55    for ( int i = 0 ; i < siz ; i++)
 56    {
 57	int v = edge[u][i].fst;
 58	if (v==pre) continue;
 59	dfs(v,dep+1,d+edge[u][i].sec,u);
 60
 61	p++;
 62	E[p] = u;
 63	depth[p] = dep;
 64    }
 65}
 66
 67
 68
 69int _min( int l,int r)
 70{
 71    if (depth[l]<depth[r]) return l;
 72    return r;
 73}
 74void rmq_init()
 75{
 76    for ( int i = 1 ; i <= 2*n+2 ; i++) dp[i][0] = i;
 77
 78    for ( int j = 1 ; (1<<j) <= 2*n+2 ; j++)
 79	for ( int i = 1 ; i + (1<<j)-1 <= 2*n+2 ; i++)
 80	dp[i][j] = _min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
 81}
 82
 83int rmq_min( int l,int r)
 84{
 85    if (l>r) swap(l,r);
 86    int k = 0 ;
 87    while (1<<(k+1)<=r-l+1) k++;
 88    return _min(dp[l][k],dp[r-(1<<k)+1][k]);
 89}
 90int solve (int x,int y)
 91{
 92    int LCA = E[rmq_min(R[x],R[y])];
 93    int res = dis[x] + dis[y] - 2 * dis[LCA];
 94    return res;
 95}
 96int main()
 97{
 98    #ifndef  ONLINE_JUDGE 
 99	freopen("code/in.txt","r",stdin);
100  #endif
101
102	ms(in,0);
103	bool ok = false;
104	while (~scanf("%d",&n)){
105	    if (ok) puts("");
106	    ok = true;
107	    for ( int i = 0 ; i <= n ; i++) edge[i].clear();
108	    for ( int i = 1 ; i <= n-1 ; i++)
109	    {
110		int u,v,w;
111		scanf("%d%d%d",&u,&v,&w);
112		edge[u].push_back(make_pair(v,w));
113		edge[v].push_back(make_pair(u,w));
114	    }
115
116
117	    p = 0 ;
118	    dfs(0,0,0,-1);
119	    rmq_init();
120
121	    scanf("%d",&q);
122	    while (q--)
123	    {
124		int x,y,z;
125		scanf("%d%d%d",&x,&y,&z);
126		int ans = solve(x,y)+solve(x,z)+solve(y,z);
127		printf("%d\n",ans/2);
128	    }
129	}
130
131#ifndef ONLINE_JUDGE  
132	fclose(stdin);
133#endif
134    return 0;
135}