一、题目描述

Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason of them is the poor design of the roads distribution, and he want to change this situation.

In order to achieve this project, he divide the city up to N regions which can be viewed as separate points. He thinks that the best design is the one that connect all region with shortest road, and he is asking you to check some of his designs.

Now, he gives you an acyclic graph representing his road design, you need to find out the shortest path to connect some group of three regions.

Input

The input contains multiple test cases! In each case, the first line contian a interger N (1 < N < 50000), indicating the number of regions, which are indexed from 0 to N-1. In each of the following N-1 lines, there are three interger Ai, Bi, Li (1 < Li < 100) indicating there's a road with length Li between region Ai and region Bi. Then an interger Q (1 < Q < 70000), the number of group of regions you need to check. Then in each of the following Q lines, there are three interger Xi, Yi, Zi, indicating the indices of the three regions to be checked.

Process to the end of file.

Output

Q lines for each test case. In each line output an interger indicating the minimum length of path to connect the three regions.

Output a blank line between each test cases.

Sample Input

4
0 1 1
0 2 1
0 3 1
2
1 2 3
0 1 2
5
0 1 1
0 2 1
1 3 1
1 4 1
2
0 1 2
1 0 3

Sample Output

3
2
2
2

二、分析
      用RMQ解决的LCA问题，详细算法：LCA问题。
三、代码

  1#include<iostream>
  2#include<cmath>
  3#include<list>
  4using namespace std;
  5int n, q;
  6struct node
  7{
  8    int lab, dis;
  9    void init(int l, int d)
 10    {
 11        lab = l; dis = d;
 12    }
 13};
 14int v1, v2, v3, len;
 15list<node> g[50001];
 16int ei, e[100002], r[50001], l[100002], d[50001];
 17bool visit[50001];
 18int pow2[18];
 19int mmin[18][100002];
 20void dfs(int u, int dep)
 21{
 22    e[++ei] = u; l[ei] = dep;
 23    if(visit[u]) return;
 24    visit[u] = true;
 25    list<node>::iterator it = g[u].begin();
 26    while(it != g[u].end())
 27    {
 28        int v = it->lab, len = it->dis;
 29        if(!visit[v])
 30        {
 31            d[v] = min(d[v], d[u] + len);
 32            dfs(v, dep+1);
 33            e[++ei] = u; l[ei] = dep;
 34            
 35        }
 36        it++;
 37    }
 38}
 39void init_rmq()
 40{
 41    ei = 0;
 42    memset(visit, 0, sizeof(visit));
 43    d[0] = 0;
 44    dfs(0, 1);
 45    memset(r, -1, sizeof(r));
 46    for(int i=1; i<=ei; i++)
 47        if(r[e[i]] == -1)
 48            r[e[i]] = i;
 49    memset(mmin, 0, sizeof(mmin));
 50    for(int i=1; i<=ei; i++)
 51        mmin[0][i] = i;
 52    int t1 = (int)(log((double)ei) / log(2.0));
 53    for(int i=1; i<=t1; i++)
 54        for(int j=1; j + pow2[i] - 1<=ei; j++)
 55        {
 56            int a = mmin[i-1][j], b = mmin[i-1][j+pow2[i-1]];
 57            if(l[a] <= l[b])
 58                mmin[i][j] = a;
 59            else
 60                mmin[i][j] = b;
 61        }
 62}
 63int rmq(int u, int v)
 64{
 65    int i = r[u], j = r[v];
 66    if(i > j) swap(i, j);
 67    int t1 = (int)(log((double)j - i + 1) / log(2.0));
 68    int a = mmin[t1][i], b = mmin[t1][j - pow2[t1] + 1];
 69    if(l[a] <= l[b])
 70        return e[a];
 71    else
 72        return e[b];
 73}
 74int main()
 75{
 76    for(int i=0; i<18; i++)
 77        pow2[i] = 1 << i;
 78    bool flag = false;
 79    while(scanf("%d", &n) != EOF)
 80    {
 81        if(flag) printf("\n");
 82        flag = true;
 83        for(int i=0; i<n; i++)
 84        {
 85            g[i].clear();
 86            d[i] = INT_MAX;
 87        }
 88        for(int i=0; i<n-1; i++)
 89        {
 90            scanf("%d%d%d", &v1, &v2, &len);
 91            node n1; n1.init(v2, len);
 92            g[v1].push_back(n1);
 93            node n2; n2.init(v1, len);
 94            g[v2].push_back(n2);
 95        }
 96        init_rmq();
 97        scanf("%d", &q);
 98        while(q--)
 99        {
100            int res = INT_MAX;
101            scanf("%d%d%d", &v1, &v2, &v3);
102            int temp = 0;
103            int lca1 = rmq(v1, v2);
104            temp = d[v1] + d[v2] - 2*d[lca1];
105            int lca2 = rmq(lca1, v3);
106            temp += d[lca1] + d[v3] - 2*d[lca2];
107            res = min(res, temp);
108            temp = 0;
109            lca1 = rmq(v1, v3);
110            temp = d[v1] + d[v3] - 2*d[lca1];
111            lca2 = rmq(lca1, v2);
112            temp += d[v2] + d[lca1] - 2*d[lca2];
113            res = min(res, temp);
114            temp = 0;
115            lca1 = rmq(v2, v3);
116            temp = d[v2] + d[v3] - 2*d[lca1];
117            lca2 = rmq(lca1, v1);
118            temp += d[v1] + d[lca1] - 2*d[lca2];
119            res = min(res, temp);
120            printf("%d\n", res);
121        }
122    }
123}