[USACO08OCT]牧场散步Pasture Walking

                                   [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.

 

树剖LCA模板

 

  1 #include <cstdio>  2 #include <cctype>  3 #include <queue>  4   5 const int MAXN=1010;  6 const int INF=0x3f3f3f3f;  7   8 int n,q,inr;  9  10 int dep[MAXN],fa[MAXN],son[MAXN],siz[MAXN],id[MAXN],top[MAXN],dis[MAXN]; 11  12 bool vis[MAXN]; 13  14 struct node { 15     int to; 16     int next; 17     int val; 18     node() {}; 19     node(int to,int val,int next):to(to),val(val),next(next) {} 20 }; 21 node e[MAXN<<1]; 22  23 int head[MAXN],tot; 24  25 inline void read(int&x) { 26     int f=1;register char c=getchar(); 27     for(x=0;!isdigit(c);c=='-'&&(f=-1),c=getchar()); 28     for(;isdigit(c);x=x*10+c-48,c=getchar()); 29     x=x*f; 30 } 31  32 inline void add(int x,int y,int val) { 33     e[++tot]=node(y,val,head[x]); 34     head[x]=tot; 35     e[++tot]=node(x,val,head[y]); 36     head[y]=tot; 37 } 38  39 void SPFA() { 40     std::queue<int> q; 41     for(int i=1;i<=n;++i) vis[i]=false,dis[i]=INF,son[i]=-1; 42     dis[1]=0; 43     q.push(1); 44     while(!q.empty()) { 45         int u=q.front(); 46         q.pop(); 47         for(int i=head[u];i;i=e[i].next) { 48             int v=e[i].to; 49             if(dis[v]>dis[u]+e[i].val) { 50                 dis[v]=dis[u]+e[i].val; 51                 if(!vis[v]) q.push(v),vis[v]=true; 52             } 53         } 54     } 55 } 56  57 void DFS_1(int u,int f) { 58     dep[u]=dep[f]+1; 59     fa[u]=f; 60     siz[u]=1; 61     for(int i=head[u];i;i=e[i].next) { 62         int v=e[i].to; 63         if(v==f) continue; 64         DFS_1(v,u); 65         siz[u]+=siz[v]; 66         if(son[u]==-1||siz[son[u]]<siz[v]) son[u]=v; 67     } 68 } 69  70 void DFS_2(int u,int tp) { 71     top[u]=tp; 72     id[u]=++inr; 73     if(son[u]==-1) return; 74     DFS_2(son[u],tp); 75     for(int i=head[u];i;i=e[i].next) { 76         int v=e[i].to; 77         if(v==fa[u]||v==son[u]) continue; 78         DFS_2(v,v); 79     } 80 } 81  82 inline int LCA(int x,int y) { 83     while(top[x]!=top[y]) { 84         if(dep[top[x]]<dep[top[y]]) x^=y^=x^=y; 85         x=fa[top[x]]; 86     } 87     if(dep[x]>dep[y]) x^=y^=x^=y; 88     return x; 89 } 90  91 int hh() { 92     read(n);read(q); 93     for(int x,y,z,i=1;i<n;++i) { 94         read(x);read(y);read(z); 95         add(x,y,z); 96     } 97     SPFA(); 98     DFS_1(1,0); 99     DFS_2(1,1);100     for(int x,y,i=1;i<=q;++i) {101         read(x);read(y);102         int lca=LCA(x,y);103         printf("%d\n",dis[x]+dis[y]-2*dis[lca]);104     }105     return 0;106 }107 108 int sb=hh();109 int main(int argc,char**argv) {;}

代码