洛谷——P2912 [USACO08OCT]牧场散步Pasture Walking(lca)
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.
代码:
#include<vector> #include<stdio.h> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define N 10001 using namespace std; vector<pair<int,int> >vec[N]; int n,m,x,y,z,fa[N],deep[N],dis[N],size[N],top[N]; int lca(int x,int y) { while(top[x]!=top[y]) { if(deep[x]<deep[y]) swap(x,y); x=fa[top[x]]; } if(deep[x]>deep[y]) swap(x,y); return x; } int dfs(int x) { size[x]=1; deep[x]=deep[fa[x]]+1; for(int i=0;i<vec[x].size();i++) { if(fa[x]!=vec[x][i].first) { fa[vec[x][i].first]=x; dis[vec[x][i].first]=dis[x]+vec[x][i].second; dfs(vec[x][i].first); size[x]+=size[vec[x][i].first]; } } } int dfs1(int x) { int t=0; if(!top[x]) top[x]=x; for(int i=0;i<vec[x].size();i++) if(vec[x][i].first!=fa[x]&&size[x]<size[vec[x][i].first]) t=vec[x][i].first; if(t) top[t]=top[x],dfs1(t); for(int i=0;i<vec[x].size();i++) if(vec[x][i].first!=fa[x]&&vec[x][i].first!=t) dfs1(vec[x][i].first); } int main() { scanf("%d%d",&n,&m); for(int i=1;i<n;i++) { scanf("%d%d%d",&x,&y,&z); vec[x].push_back(make_pair(y,z)); vec[y].push_back(make_pair(x,z)); } dfs(1); dfs1(1); for(int i=1;i<=m;i++) { scanf("%d%d",&x,&y); printf("%d\n",dis[x]+dis[y]-2*dis[lca(x,y)]); } return 0; }
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