lca倍增dp

题目链接：题目

题意：在一个树上，选取两个点，是他们的差值最大，不过较大值应该在较小值路径的前面。

解题思路：倍增lca，然后采用倍增维护信息，第一次学倍增dp。

代码：

/* ***********************************************Author :xianxingwuguanCreated Time :2014/1/16 20:42:49File Name :C.cpp************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;const int maxn=60050;const int inf=100010000;const int DEG=20;int head[maxn],tol,mx[maxn][DEG],mi[maxn][DEG],dp[maxn][DEG],dp2[maxn][DEG],fa[maxn][DEG],dep[maxn];int weight[maxn];struct node{      int next,to;}edge[3*maxn];void add(int u,int v){      edge[tol].to=v;      edge[tol].next=head[u];      head[u]=tol++;}int n;void dfs(int u,int pre){      dep[u]=dep[pre]+1;      for(int i=head[u];i!=-1;i=edge[i].next)      {     int v=edge[i].to;     if(v==pre)continue;     fa[v][0]=u;     mx[v][0]=max(weight[u],weight[v]);     mi[v][0]=min(weight[u],weight[v]);     dp[v][0]=weight[u]-weight[v];     dp2[v][0]=weight[v]-weight[u];     for(int j=1;j<20;j++)     {    fa[v][j]=fa[fa[v][j-1]][j-1];    mx[v][j]=max(mx[v][j-1],mx[fa[v][j-1]][j-1]);    mi[v][j]=min(mi[v][j-1],mi[fa[v][j-1]][j-1]);    dp[v][j]=max(dp[v][j-1],dp[fa[v][j-1]][j-1]);    dp[v][j]=max(dp[v][j],mx[fa[v][j-1]][j-1]-mi[v][j-1]);    dp2[v][j]=max(dp2[v][j-1],dp2[fa[v][j-1]][j-1]);    dp2[v][j]=max(dp2[v][j],mx[v][j-1]-mi[fa[v][j-1]][j-1]);     }     dfs(v,u);      }}int lca(int x,int y){      if(dep[x]<dep[y])swap(x,y);      for(int i=0;i<20;i++)if((dep[x]-dep[y])&(1<<i))x=fa[x][i];      if(x==y)return x;      for(int i=19;i>=0;i--)if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];      return fa[x][0];}void init(){      fill(mx[0],mx[n+1],-inf);      fill(mi[0],mi[n+1],inf);      fill(dp[0],dp[n+1],-inf);      fill(dp2[0],dp2[n+1],-inf);      dep[0]=0;      dfs(1,0);}int getmin(int u,int lca,int dp[][DEG]) {      int ans=inf;      int del=dep[u] - dep[lca];      for(int i=DEG-1;i>=0;i--)     if(del & (1<<i))     {          ans=min(ans,dp[u][i]);          u=fa[u][i];      }      return ans;  }  int getmax(int x,int lca,int dp[][DEG]){      int ans=0;      int del=dep[x]-dep[lca];      for(int i=DEG-1;i>=0;i--)     if(del & (1<<i))    {          ans=max(ans,dp[x][i]);          x=fa[x][i];      }      return ans;  }  int gao(int x,int lca,int dp[][DEG]) {      int ans=0,tmp=0;      int del=dep[x]-dep[lca];      for(int i=0;i<DEG;i++)     if(del & (1<<i))     {          ans=max(ans,dp[x][i]);          ans=max(ans,tmp-mi[x][i]);          tmp=max(tmp,mx[x][i]);          x=fa[x][i];      }      return ans;  }  int gao2(int x,int lca,int dp[][DEG]) {          int ans=0,tmp=inf;      int del=dep[x]-dep[lca];      for(int i=DEG-1;i>=0;i--) if(del & (1<<i))     {          ans=max(ans,dp[x][i]);          ans=max(ans,-(tmp-mx[x][i]));          tmp=min(tmp,mi[x][i]);          x=fa[x][i];      }      return ans;  }  int solve(int x,int y){      int p=lca(x,y);      int a=gao2(x,p,dp);      int b=gao(y,p,dp2);      int c=getmin(x,p,mi);      int d=getmax(y,p,mx);      int ans=max(max(a,b),d-c);      return  ans;}int main(){      //freopen("data.in","r",stdin);      //freopen("data.out","w",stdout);      int i,j,k,m;      while(~scanf("%d",&n))      {            memset(head,-1,sizeof(head));tol=0;     for(i=1;i<=n;i++)scanf("%d",&weight[i]);     for(i=1;i<n;i++)     {    scanf("%d%d",&j,&k);    add(j,k);    add(k,j);     }            init();     int Q;     scanf("%d",&Q);     while(Q--)     {    scanf("%d%d",&i,&j);    printf("%d\n",solve(i,j));     }      }      return 0;}