洛谷 P2912 [USACO08OCT]牧场散步Pasture Walking
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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.
LCA裸题
#include <ctype.h> #include <cstdio> #define N 1005 void read(int &x) { x=0;register char ch=getchar(); for(;!isdigit(ch);ch=getchar()); for(;isdigit(ch);ch=getchar()) x=x*10+ch-‘0‘; } struct Edge { int next,to,dis; Edge (int next=0,int to=0,int dis=0) :next(next),to(to),dis(dis){} }edge[N<<1]; int dis[N],dad[N][25],dep[N],head[N],cnt,n,q; void insert(int u,int v,int w) { edge[++cnt]=Edge(head[u],v,w); head[u]=cnt; } void swap(int &x,int &y) { int tmp=y; y=x; x=tmp; } void dfs(int x) { dep[x]=dep[dad[x][0]]+1; for(int i=0;dad[x][i];i++) dad[x][i+1]=dad[dad[x][i]][i]; for(int u=head[x];u;u=edge[u].next) { int v=edge[u].to; if(dad[x][0]!=v) { dad[v][0]=x; dis[v]=dis[x]+edge[u].dis; dfs(v); } } } int lca(int x,int y) { if(dep[x]>dep[y]) swap(x,y); for(int i=20;i>=0;i--) if(dep[dad[y][i]]>=dep[x]) y=dad[y][i]; if(x==y) return x; for(int i=20;i>=0;i--) if(dad[y][i]!=dad[x][i]) y=dad[y][i],x=dad[x][i]; return dad[x][0]; } int main() { read(n); read(q); for(int x,y,z,i=1;i<n;i++) { read(x); read(y); read(z); insert(x,y,z); insert(y,x,z); } dfs(1); for(int x,y;q--;) { read(x); read(y); int LCA=lca(x,y); printf("%d\n",dis[x]+dis[y]-2*dis[LCA]); } return 0; }
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