洛谷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

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.

典型的LCA+暴力。。。

LCA首选倍增。。。

附代码：

#include<iostream>#include<algorithm>#include<cstdio>#define MAXN 1010#define MAX 999999999using namespace std;int n,m,c=1;int a[MAXN][MAXN],f[MAXN][20],deep[MAXN];inline int read(){int date=0,w=1;char c=0;while(c<'0'||c>'9'){if(c=='-')w=-1;c=getchar();}while(c>='0'&&c<='9'){date=date*10+c-'0';c=getchar();}return date*w;}void buildtree(int rt){for(int i=1;i<=n;i++){if(!deep[i]&&rt!=i&&a[rt][i]!=MAX){deep[i]=deep[rt]+1;f[i][0]=rt;buildtree(i);}}}void step(){for(int i=1;i<=19;i++)for(int j=1;j<=n;j++)f[j][i]=f[f[j][i-1]][i-1];}int LCA(int x,int y){if(deep[x]<deep[y])swap(x,y);for(int i=19;i>=0;i--)if(deep[f[x][i]]>=deep[y])x=f[x][i];if(x==y)return x;for(int i=19;i>=0;i--)if(f[x][i]!=f[y][i]){x=f[x][i];y=f[y][i];}return f[x][0];}void work(int x,int y){int fa=LCA(x,y),s1=0,s2=0;for(int i=x;i!=fa;i=f[i][0])s1+=a[i][f[i][0]];for(int i=y;i!=fa;i=f[i][0])s2+=a[i][f[i][0]];printf("%d\n",s1+s2);}int main(){int u,v,w;n=read();m=read();for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)a[i][j]=(i==j)?0:MAX;for(int i=1;i<n;i++){u=read();v=read();w=read();if(a[u][v]>w)a[u][v]=a[v][u]=w;}deep[1]=1;buildtree(1);step();while(m--){u=read();v=read();work(u,v);}return 0;}