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
4 2
2 1 2
4 3 2
1 4 3
1 2
3 2
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 ;
}