Fzu Problem 2082 过路费 LCT,动态树

题目：http://acm.fzu.edu.cn/problem.php?pid=2082

Problem 2082 过路费

Accept: 528    Submit: 1654
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

有n座城市，由n-1条路相连通，使得任意两座城市之间可达。每条路有过路费，要交过路费才能通过。每条路的过路费经常会更新，现问你，当前情况下，从城市a到城市b最少要花多少过路费。

 Input

有多组样例，每组样例第一行输入两个正整数n,m(2 <= n<=50000，1<=m <= 50000),接下来n-1行，每行3个正整数a b c，(1 <= a,b <= n , a != b , 1 <= c <= 1000000000).数据保证给的路使得任意两座城市互相可达。接下来输入m行，表示m个操作，操作有两种：一. 0 a b，表示更新第a条路的过路费为b，1 <= a <= n-1 ； 二. 1 a b ， 表示询问a到b最少要花多少过路费。

 Output

对于每个询问，输出一行，表示最少要花的过路费。

 Sample Input

2 3
1 2 1
1 1 2
0 1 2
1 2 1

 Sample Output

1
2

 Source

FOJ有奖月赛-2012年4月（校赛热身赛）

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题解：

直接动态树维护树上两点间的和即可。。。

注意开long long。。。

 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 #define LL long long
 #define MAXN 50010
 struct node
 {
     int left,right,val;
     LL sum;
 }tree[MAXN*];
 int father[MAXN*],Stack[*MAXN],rev[*MAXN];
 int read()
 {
     int s=,fh=;char ch=getchar();
     while(ch<''||ch>''){if(fh=='-')fh=-;ch=getchar();}
     while(ch>=''&&ch<=''){s=s*+(ch-'');ch=getchar();}
     return s*fh;
 }
 int isroot(int x)
 {
     return tree[father[x]].left!=x&&tree[father[x]].right!=x;
 }
 void pushdown(int x)
 {
     int l=tree[x].left,r=tree[x].right;
     if(rev[x]!=)
     {
         rev[x]^=;rev[l]^=;rev[r]^=;
         swap(tree[x].left,tree[x].right);
     }
 }
 void Pushup(int x)
 {
     int l=tree[x].left,r=tree[x].right;
     tree[x].sum=tree[l].sum+tree[r].sum+tree[x].val;
 }
 void rotate(int x)
 {
     int y=father[x],z=father[y];
     if(!isroot(y))
     {
         if(tree[z].left==y)tree[z].left=x;
         else tree[z].right=x;
     }
     if(tree[y].left==x)
     {
         father[x]=z;father[y]=x;tree[y].left=tree[x].right;tree[x].right=y;father[tree[y].left]=y;
     }
     else
     {
         father[x]=z;father[y]=x;tree[y].right=tree[x].left;tree[x].left=y;father[tree[y].right]=y;
     }
     Pushup(y);Pushup(x);
 }
 void splay(int x)
 {
     int top=,i,y,z;Stack[++top]=x;
     for(i=x;!isroot(i);i=father[i])Stack[++top]=father[i];
     for(i=top;i>=;i--)pushdown(Stack[i]);
     while(!isroot(x))
     {
         y=father[x];z=father[y];
         if(!isroot(y))
         {
             if((tree[y].left==x)^(tree[z].left==y))rotate(x);
             else rotate(y);
         }
         rotate(x);
     }
 }
 void access(int x)
 {
     int last=;
     while(x!=)
     {
         splay(x);
         tree[x].right=last;Pushup(x);
         last=x;x=father[x];
     }
 }
 void makeroot(int x)
 {
     access(x);splay(x);rev[x]^=;
 }
 void link(int u,int v)
 {
     makeroot(u);father[u]=v;splay(u);
 }
 void cut(int u,int v)
 {
     makeroot(u);access(v);splay(v);father[u]=tree[v].left=;
 }
 int findroot(int x)
 {
     access(x);splay(x);
     while(tree[x].left!=)x=tree[x].left;
     return x;
 }
 int main()
 {
     int n,m,i,fh,aa,bb,cc;
     while(scanf("%d %d",&n,&m)!=EOF)
     {
         for(i=;i<=*n;i++)tree[i].sum=tree[i].left=tree[i].right=tree[i].val=father[i]=rev[i]=;
         for(i=;i<n;i++)
         {
             aa=read();bb=read();cc=read();
             tree[n+i].val=cc;
             link(aa,n+i);link(n+i,bb);
         }
         for(i=;i<=m;i++)
         {
             fh=read();aa=read();bb=read();
             if(fh==)
             {
                 splay(n+aa);
                 tree[n+aa].val=bb;
             }
             else
             {
                 makeroot(aa);access(bb);splay(bb);
                 printf("%lld\n",tree[bb].sum);
             }
         }
     }
     return ;
 }