洛谷P2912 [USACO08OCT]牧场散步Pasture Walking [2017年7月计划 树上问题 01]

P2912 [USACO08OCT]牧场散步Pasture Walking

题目描述

The N cows (2 <= N <= 1,000) conveniently numbered 1..N are grazing among the N pastures also conveniently numbered 1..N. Most conveniently of all, cow i is grazing in pasture i.

Some pairs of pastures are connected by one of N-1 bidirectional walkways that the cows can traverse. Walkway i connects pastures A_i and B_i (1 <= A_i <= N; 1 <= B_i <= N) and has a length of L_i (1 <= L_i <= 10,000).

The walkways are set up in such a way that between any two distinct pastures, there is exactly one path of walkways that travels between them. Thus, the walkways form a tree.

The cows are very social and wish to visit each other often. Ever in a hurry, they want you to help them schedule their visits by computing the lengths of the paths between 1 <= L_i <= 10,000 pairs of pastures (each pair given as a query p1,p2 (1 <= p1 <= N; 1 <= p2 <= N).

POINTS: 200

有N(2<=N<=1000)头奶牛，编号为1到W，它们正在同样编号为1到N的牧场上行走.为了方 便，我们假设编号为i的牛恰好在第i号牧场上.

有一些牧场间每两个牧场用一条双向道路相连，道路总共有N - 1条，奶牛可以在这些道路 上行走.第i条道路把第Ai个牧场和第Bi个牧场连了起来(1 <= A_i <= N; 1 <= B_i <= N)，而它的长度 是 1 <= L_i <= 10,000.在任意两个牧场间，有且仅有一条由若干道路组成的路径相连.也就是说，所有的道路构成了一棵树.

奶牛们十分希望经常互相见面.它们十分着急，所以希望你帮助它们计划它们的行程，你只 需要计算出Q(1 < Q < 1000)对点之间的路径长度•每对点以一个询问p1,p2 (1 <= p1 <= N; 1 <= p2 <= N). 的形式给出.

输入输出格式

输入格式：

• Line 1: Two space-separated integers: N and Q

• Lines 2..N: Line i+1 contains three space-separated integers: A_i, B_i, and L_i

• Lines N+1..N+Q: Each line contains two space-separated integers
  representing two distinct pastures between which the cows wish to
  travel: p1 and p2

输出格式：

• Lines 1..Q: Line i contains the length of the path between the two pastures in query i.

输入输出样例

输入样例#1：

4 2
2 1 2
4 3 2
1 4 3
1 2
3 2

输出样例#1：

2
7

说明

Query 1: The walkway between pastures 1 and 2 has length 2.

Query 2: Travel through the walkway between pastures 3 and 4, then the one between 4 and 1, and finally the one between 1 and 2, for a total length of 7.

裸LCA,预处理的时候顺带求到根节点的路径长，答案为len[u] + len[v] - 2 * len[lca(va, vb）]

 #include <bits/stdc++.h>

 const int MAXN =  + ;

 inline void read(int &x)
 {
     x = ;char ch = getchar();char c = ch;
     while(ch > '' || ch < '')c = ch, ch = getchar();
     while(ch <= '' && ch >= '')x = x *  + ch - '', ch = getchar();
     if(c == '-')x = -x;
 }

 inline void swap(int &a, int &b)
 {
     int tmp = a;
     a = b;
     b = tmp;
 }

 struct Edge{int u,v,w,next;}edge[MAXN >> ];
 int head[MAXN],root,cnt,deep[MAXN],len[MAXN],p[MAXN][];
 int n,m,tmp1,tmp2,tmp3;
 bool b[MAXN];

 inline void insert(int a, int b, int c)
 {
     edge[++cnt] = {a,b,c,head[a]};
     head[a] = cnt;
 }

 void dfs(int u, int step)
 {
     for(int pos = head[u];pos;pos = edge[pos].next)
     {
         int v = edge[pos].v;
         if(!b[v])
         {
             b[v] = true;
             deep[v] = step + ;
             len[v] = len[u] + edge[pos].w;
             p[v][] = u;
             dfs(v, step + );
         }
     }
 }

 inline void yuchuli()
 {
     b[root] = true;
     len[root] = ;
     deep[root] = ;
     dfs(root, );
     for(int i = ;( << i) <= n;i ++)
     {
         for(int j = ;j <= n;j ++)
         {
             p[j][i] = p[p[j][i - ]][i - ];
         }
     }
 }

 inline int lca(int va, int vb)
 {
     if(deep[va] < deep[vb])swap(va, vb);
     int M = ;
     for(;( << M) <= n;M ++);
     M --;
     for(int i = M;i >= ;i --)
         if(deep[va] - ( << i) >= deep[vb])
             va = p[va][i];
     if(va == vb)return va;
     for(int i = M;i >= ;i --)
         if(p[va][i] != p[vb][i])
         {
             va = p[va][i];
             vb = p[vb][i];
         }
     return p[va][];
 }

 int main()
 {
     read(n);read(m);
     for(int i = ;i < n;i ++)
     {
         read(tmp1);read(tmp2);read(tmp3);
         insert(tmp1, tmp2, tmp3);
         insert(tmp2, tmp1, tmp3);
     }
     root = ;
     yuchuli();
     for(int i = ;i <= m;i ++)
     {
         read(tmp1);read(tmp2);
         printf("%d\n", len[tmp1] + len[tmp2] -  * len[lca(tmp1, tmp2)]);
     }
     return ;
 }