[USACO08OCT]牧场散步Pasture Walking

Posted 2020-10-06 心之所向 素履以往

                                   [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