Housewife Wind

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 11250		Accepted: 3111

After their royal wedding, Jiajia and Wind hid away in XX Village, to enjoy their ordinary happy life. People in XX Village lived in beautiful huts. There are some pairs of huts connected by bidirectional roads. We say that huts in the same pair directly connected. XX Village is so special that we can reach any other huts starting from an arbitrary hut. If each road cannot be walked along twice, then the route between every pair is unique. 

Since Jiajia earned enough money, Wind became a housewife. Their children loved to go to other kids, then make a simple call to Wind: ‘Mummy, take me home!‘ 

At different times, the time needed to walk along a road may be different. For example, Wind takes 5 minutes on a road normally, but may take 10 minutes if there is a lovely little dog to play with, or take 3 minutes if there is some unknown strange smell surrounding the road. 

Wind loves her children, so she would like to tell her children the exact time she will spend on the roads. Can you help her? 

The first line contains three integers n, q, s. There are n huts in XX Village, q messages to process, and Wind is currently in hut s. n < 100001 , q < 100001. 

The following n-1 lines each contains three integers a, b and w. That means there is a road directly connecting hut a and b, time required is w. 1<=w<= 10000. 

The following q lines each is one of the following two types: 

Message A: 0 u 
A kid in hut u calls Wind. She should go to hut u from her current position. 
Message B: 1 i w 
The time required for i-th road is changed to w. Note that the time change will not happen when Wind is on her way. The changed can only happen when Wind is staying somewhere, waiting to take the next kid. 

For each message A, print an integer X, the time required to take the next child.

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

1
3


这个题首先很容易想到是LCA，题中的每次修改可以惊天暴力地每次修改后更新一次w[]数组，复杂度本蒟蒻不会计算，猜测极限数据60s左右能跑出来吧......
那么显然这个做法是不合理的，此时需要知道一个叫做dfs序（或者叫欧拉序）的事情。
dfs序是一棵树在进行dfs遍历是组成的结点序列，是序列，不是数列。
dfs序有两种记法，本题用到的记法是每个结点进出都加进序列，显然发现每个结点都恰出现了两次，显然此时可以发现题中每次的修改仅影响到了dfs序中以某同一结点为两端的一个闭区间，这句话好绕口啊。
当然此题是个树剖水题，spli奆奆说的。


贴代码，线段树注意变通一下
代码丑陋勿喷！

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 using namespace std;
  6 int n,q,s,cnt,tot;
  7 struct data{
  8     int next,to,dis;
  9 }edge[400010];
 10 int head[200010],getin[200010],getout[200010],f[200010][25],deep[200010],w[200010],tree[800100];
 11 void add(int start,int end,int ds){
 12     edge[++cnt].next=head[start];    
 13     edge[cnt].to=end;
 14     edge[cnt].dis=ds;
 15     head[start]=cnt;
 16 }
 17 void dfs(int u){
 18     getin[u]=++tot;
 19     for(int i=head[u];i;i=edge[i].next)
 20         if(!deep[edge[i].to]){
 21             deep[edge[i].to]=deep[u]+1;
 22             w[edge[i].to]=w[u]+edge[i].dis;
 23             f[edge[i].to][0]=u;
 24             dfs(edge[i].to);
 25         }
 26     getout[u]=++tot;
 27 }
 28 void work(){
 29     for(int j=0;(1<<j)<=n;j++)
 30         for(int i=1;i<=n;i++)
 31             if(f[i][j-1]!=-1) f[i][j]=f[f[i][j-1]][j-1];
 32 }
 33 int lca(int uu,int vv){
 34     if(deep[uu]<deep[vv]) swap(uu,vv);
 35     int i=0;
 36     for(i=0;(1<<i)<=n;i++);
 37     i--;
 38     for(int j=i;j>=0;j--)
 39         if(deep[uu]-(1<<j)>=deep[vv]) uu=f[uu][j];
 40     if(uu==vv) return uu;
 41     for(int j=i;j>=0;j--)
 42         if(f[uu][j]!=-1&&f[uu][j]!=f[vv][j]){
 43             uu=f[uu][j];
 44             vv=f[vv][j];
 45         }
 46     return f[uu][0];
 47 }
 48 void build(int pos,int ll,int rr){
 49     tree[pos]=0;
 50     if(ll==rr) return;
 51     int mid=(ll+rr)>>1;
 52     build(pos<<1,ll,mid);
 53     build(pos<<1|1,mid+1,rr);
 54     return;
 55 }
 56 void down(int pos){
 57     if(tree[pos]){
 58         tree[pos<<1]+=tree[pos];
 59         tree[pos<<1|1]+=tree[pos];
 60         tree[pos]=0;
 61     }
 62     return;
 63 }
 64 int sum(int xx,int yy,int pos,int ll,int rr){
 65     if(xx==ll&&yy==rr) return tree[pos];
 66     down(pos);
 67     int mid=(ll+rr)>>1;
 68     if(yy<=mid) return sum(xx,yy,pos<<1,ll,mid);
 69     else if(xx>mid) return sum(xx,yy,pos<<1|1,mid+1,rr);
 70         else return sum(xx,mid,pos<<1,ll,mid)+sum(mid+1,yy,pos<<1|1,mid+1,rr);
 71 }
 72 void updata(int xx,int yy,int pos,int ll,int rr,int dd){
 73     if(xx==ll&&yy==rr){
 74         tree[pos]+=dd;
 75         return;
 76     }
 77     down(pos);
 78     int mid=(ll+rr)>>1;
 79     if(yy<=mid) updata(xx,yy,pos<<1,ll,mid,dd);
 80     else if(xx>mid) updata(xx,yy,pos<<1|1,mid+1,rr,dd);
 81         else{
 82             updata(xx,mid,pos<<1,ll,mid,dd);
 83             updata(mid+1,yy,pos<<1|1,mid+1,rr,dd);
 84         }
 85     return;
 86 }
 87 int main(){
 88     scanf("%d%d%d",&n,&q,&s);
 89     int u,v,d;
 90     for(int i=1;i<n;i++){
 91         scanf("%d%d%d",&u,&v,&d);
 92         add(u,v,d);
 93         add(v,u,d);
 94     }
 95     memset(f,-1,sizeof(f));
 96     deep[1]=1;
 97     dfs(1);
 98     work();
 99     int od,nm,vl,e;
100     build(1,1,2*n);
101     for(int i=1;i<=q;i++){
102         scanf("%d",&od);
103         if(od){
104             scanf("%d%d",&nm,&vl);
105             int tmp,tu,tv;
106             tu=edge[nm<<1].to;
107             tv=edge[(nm<<1)-1].to;
108             if(f[tu][0]==tv) tmp=tu;
109             else tmp=tv;
110             updata(getin[tmp],getout[tmp],1,1,2*n,vl-edge[nm<<1].dis);
111             edge[nm<<1].dis=vl;
112             edge[(nm<<1)-1].dis=vl;
113         }
114         if(!od){
115             scanf("%d",&e);
116             int d1=w[s]+sum(getin[s],getin[s],1,1,n<<1);
117             int d2=w[e]+sum(getin[e],getin[e],1,1,n<<1);
118             int d3=w[lca(s,e)]+sum(getin[lca(s,e)],getin[lca(s,e)],1,1,n<<1);
119             printf("%d\n",d1+d2-2*d3);
120             s=e;
121         }
122     }
123     return 0;
124 }