[BZOJ]1602: [Usaco2008 Oct]牧场行走

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题目大意:给你一棵树以及相邻节点之间的距离，之后询问2点之间的距离。
用dis数组存根节点到其他节点的距离，则有distance(a,b)=dis[a]+dis[b]-dis[lca(a,b)]*2,所以这题可以用树上倍增求LCA。

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;inline char tc(void){    static char fl[10000+1],*A=fl,*B=fl;    return A==B?B=(A=fl)+fread(fl,1,10000,stdin),A==B?EOF:*A++:*A++;}inline void read(int &a){    a=0;static char c;    while((c=tc())<'0'||c>'9');    while(c>='0'&&c<='9')        a=a*10+c-'0',c=tc();    return ;}struct Edge{    int to,cost;    Edge(int a,int b):to(a),cost(b){}};int n,q,deep[1001],x,y,z,dis[1001],fa[1001][12];vector<Edge>s[1001];void dfs(int x){    int now=1;    while(deep[x]>(1<<now))        fa[x][now]=fa[fa[x][now-1]][now-1],++now;    for (int i=0;i<s[x].size();++i)        if(!deep[s[x][i].to])            fa[s[x][i].to][0]=x,            dis[s[x][i].to]=dis[x]+s[x][i].cost,            deep[s[x][i].to]=deep[x]+1,            dfs(s[x][i].to);    return ;}int lca(int x,int y){    if(deep[x]>deep[y])        swap(x,y);    int k=deep[y]-deep[x];    for (int i=0;i<12;++i)        if(k&(1<<i))            y=fa[y][i];    if(x==y)        return x;    for (int i=11;i>=0;--i)        if(fa[x][i]!=fa[y][i])            x=fa[x][i],y=fa[y][i];    return fa[x][0];}int main(void){    register int i;    read(n),read(q);    for(i=1;i<n;++i)        read(x),read(y),read(z),s[x].push_back(Edge(y,z)),s[y].push_back(Edge(x,z));    deep[1]=1;    dfs(1);    for (i=1;i<=q;++i)        read(x),read(y),printf("%d\n",dis[x]+dis[y]-dis[lca(x,y)]*2);    return 0;}

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