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Housewife Wind

http://poj.org/problem?id=2763

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 14820		Accepted: 4097

Description

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?

Input

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.

Output

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

Sample Input

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

Sample Output

1
3

Source

POJ Monthly--2006.02.26,zgl & twb 邊權樹鏈剖分模板題 比較一條邊上哪個點的深度大，就把權值加在哪個點上

  1 #include<iostream>
  2 #include<cstring>
  3 #include<string
 
>
  4 #include<cmath>
  5 #include<cstdio>
  6 #include<algorithm>
  7 #include<vector>
  8 #define maxn 200005
  9 #define MAXN 200005
 10 #define lson l,mid,rt<<1
 11 #define rson mid+1,r,rt<<1|1
 12 using namespace std;
 13 
 14 long long tree[maxn<<3];
 15 int n;
 16 int v[maxn],val[maxn];
 17 int dep[maxn],fa[maxn],siz[maxn],son[maxn],id[maxn],top[maxn],cnt;
 18 int co,head[MAXN];
 19 struct Edge {
 20     int to, next;
 21 }edge[MAXN];
 22 struct E {
 23     int u, v, c;
 24 }e[MAXN];
 25 void addedge(int u, int v) {
 26     edge[cnt].to = v;
 27     edge[cnt].next = head[u];
 28     head[u] = cnt++;
 29 }
 30 struct sair{
 31     int x,y,len;
 32 }p[maxn];
 33 
 34 void pushup(int rt){
 35     tree[rt]=(tree[rt<<1]+tree[rt<<1|1]);
 36 }
 37 
 38 void build(int l,int r,int rt){
 39     if(l==r){
 40         tree[rt]=0;
 41         return;
 42     }
 43     int mid=(l+r)/2;
 44     build(lson);
 45     build(rson);
 46     pushup(rt);
 47 }
 48 
 49 void add(int L,int R,int k,int l,int r,int rt){
 50     if(L<=l&&R>=r){
 51         tree[rt]=k;
 52         return;
 53     }
 54     int mid=(l+r)/2;
 55     if(L<=mid) add(L,R,k,lson);
 56     if(R>mid) add(L,R,k,rson);
 57     pushup(rt);
 58 }
 59 
 60 long long query(int L,int R,int l,int r,int rt){
 61     if(L<=l&&R>=r){
 62         return tree[rt];
 63     }
 64     int mid=(l+r)/2;
 65     long long ans=0;
 66     if(L<=mid) ans+=query(L,R,lson);
 67     if(R>mid) ans+=query(L,R,rson);
 68     pushup(rt);
 69     return ans;
 70 }
 71 
 72 void dfs1(int now,int f,int deep){
 73     dep[now]=deep;
 74     siz[now]=1;
 75     fa[now]=f;
 76     int maxson=-1;
 77     for(int i=head[now];~i;i=edge[i].next){
 78         if(edge[i].to != fa[now]) {
 79             dfs1(edge[i].to,now,deep+1);
 80             siz[now]+=siz[edge[i].to];
 81             if(siz[edge[i].to]>maxson){
 82                 maxson=siz[edge[i].to];
 83                 son[now]=edge[i].to;
 84             }
 85         }
 86     }
 87 }
 88 
 89 void dfs2(int now,int topp){
 90     id[now]=++cnt;
 91     val[cnt]=v[now];
 92     top[now]=topp;
 93     if(!son[now]) return;
 94     dfs2(son[now],topp);
 95     for(int i=head[now];~i;i=edge[i].next){
 96         int vvv = edge[i].to;
 97         if(vvv==son[now]||vvv==fa[now]) continue;
 98         dfs2(vvv,vvv);
 99     }
100 }
101 
102 long long qRange(int x,int y){
103     int t1 = top[x], t2 = top[y];
104     long long res = 0;
105     while(t1 != t2) {
106         if(dep[t1] < dep[t2]) {
107             swap(t1, t2); swap(x, y);
108         }
109         res += query(id[t1], id[x], 1, n, 1);
110         x = fa[t1]; t1 = top[x];
111     }
112     if(x == y) return res;
113     if(dep[x] > dep[y]) swap(x, y);
114     return res + query(id[son[x]], id[y], 1, n, 1);
115 }
116 
117 void addRange(int x,int y,int k){
118     while(top[x]!=top[y]){
119         if(dep[top[x]]<dep[top[y]]) swap(x,y);
120         add(id[top[x]],id[x],k,1,n,1);
121         x=fa[top[x]];
122     }
123     if(dep[x]>dep[y]) swap(x,y);
124     add(id[x],id[y],k,1,n,1);
125 }
126 
127 int main(){
128     int m,r;
129     scanf("%d %d %d",&n,&m,&r);
130     memset(head, -1, sizeof head);
131     int pos,z,x,y;
132     co=0;
133     for(int i=1;i<n;i++){
134         scanf("%d %d %d",&p[i].x,&p[i].y,&p[i].len);
135         addedge(p[i].x,p[i].y);
136         addedge(p[i].y,p[i].x);
137     }
138     cnt=0;
139     int xx;
140     dfs1(1,0,1);
141     dfs2(1,1);
142     build(1,n,1);
143     for(int i=1;i<n;i++){
144         if(dep[p[i].x]<dep[p[i].y]) xx=p[i].y;
145         else xx=p[i].x;
146         addRange(xx,xx,p[i].len);
147     }
148     for(int i=1;i<=m;i++){
149         scanf("%d %d",&pos,&x);
150         if(!pos){
151             printf("%lld\n",qRange(x,r));
152             r=x;
153         }
154         else if(pos){
155             scanf("%d",&y);
156             p[x].len=y;
157             if(dep[p[x].x]<dep[p[x].y]) xx=p[x].y;
158             else xx=p[x].x;
159             addRange(xx,xx,p[x].len);
160         }
161     }
162 
163 }

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邊權樹鏈剖分