spoj375 QTREE - Query on a tree

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.

We will ask you to perfrom some instructions of the following form:

• CHANGE i ti : change the cost of the i-th edge to ti
  or
• QUERY a b : ask for the maximum edge cost on the path from node a to node b

Input

The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.

For each test case:

• In the first line there is an integer N (N <= 10000),
• In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 1000000),
• The next lines contain instructions "CHANGE i ti" or "QUERY a b",
• The end of each test case is signified by the string "DONE".

There is one blank line between successive tests.

Output

For each "QUERY" operation, write one integer representing its result.

Example

Input:
1

3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE

Output:
1
3我也不知道这个oj怎么注册，大概找了一份标程拍了一上午，应该没问题。就是一个树剖裸题，树单点修改和查询（u，v）路径最大值。

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <cstdio>
  5 const int N =  100000 + 11 , inf = 1 << 30;
  6 using namespace std ;
  7 int n, head[N] , tot = 0,sum;
  8 struct id
  9 {
 10     int fro,nxt,to,val;
 11 } E[N<<1];
 12 void add( int u , int v , int val )
 13 {
 14     E[++tot] = (id){u,head[u],v,val};head[u] = tot;
 15     E[tot+n-1] = (id){v,head[v],u,val};head[v] = tot+n-1;
 16 }
 17 struct Node
 18 {
 19     int deep,fa,size,zson,top,pos;
 20 } node[N];
 21 struct seg
 22 {
 23     int l,r,val;
 24 }T[N<<2]; int cnt;
 25
 26 void dfs1( int u, int pre, int dep )
 27 {
 28     node[u] = (Node){dep,pre,1,-1};
 29     for(int i = head[u];~i;i = E[i].nxt)
 30     {
 31         int v = E[i].to ;
 32         if( v == pre ) continue;
 33         dfs1(v,u,dep+1);
 34         node[u].size += node[v].size;
 35         if(node[u].zson == -1 || node[v].size > node[node[u].zson].size) node[u].zson = v;
 36     }
 37
 38 }
 39 void dfs2(int u ,int p)
 40 {
 41     node[u].top = p; node[u].pos = ++sum;
 42     if(~node[u].zson)dfs2(node[u].zson,node[u].top);
 43     for(int i = head[u];~i;i = E[i].nxt)
 44     {
 45         int v = E[i].to;
 46         if(v == node[u].fa || v == node[u].zson)continue;
 47         dfs2(v,v);
 48     }
 49 }
 50
 51 void Init()
 52 {
 53     scanf("%d",&n);
 54     memset(head,-1,sizeof(head));
 55     tot = 0 , sum = 0,cnt = 1;
 56     int a,b,c;
 57     for(int i = 1 ; i < n; ++i)
 58     {
 59         scanf("%d%d%d",&a,&b,&c);
 60         add(a,b,c);
 61     }
 62     dfs1(1,0,0);
 63     dfs2(1,1);
 64 }
 65
 66 void updata( int l,int r,int &num, int cp , int val )
 67 {
 68
 69     if(!num) { num = ++cnt; T[num] = (seg){0,0,-inf};}
 70     if( l == r ){ T[num].val = val ; return ;}
 71     int mid =  ( l + r ) >> 1 ;
 72     if(mid >= cp)
 73     {
 74         updata(l,mid,T[num].l,cp,val) ;
 75         T[num].val = T[T[num].l].val;
 76         if(T[num].r) T[num].val = max(T[num].val,T[T[num].r].val);
 77     }
 78     else
 79     {
 80         updata(mid+1,r,T[num].r,cp,val) ;
 81         T[num].val = T[T[num].r].val;
 82         if(T[num].l) T[num].val = max(T[num].val,T[T[num].l].val);
 83     }
 84
 85 }
 86
 87 int query( int l , int r, int num , int L , int R )
 88 {
 89     if(l == L && r == R) return T[num].val;
 90     int mid = (l + r) >> 1;
 91     if(mid >= R)return query(l,mid,T[num].l,L,R);
 92     else if(L > mid) return query(mid+1,r,T[num].r,L,R);
 93     else return max(query(l,mid,T[num].l,L,mid),query(mid+1,r,T[num].r,mid+1,R));
 94 }
 95
 96 int lca(int x,int y)
 97 {
 98     int ans = -inf;
 99     while(node[x].top != node[y].top)
100     {
101         if(node[node[x].top].deep < node[node[y].top].deep) swap(x,y);
102         ans = max(ans , query(1,n,1,node[node[x].top].pos,node[x].pos));
103         x = node[node[x].top].fa;
104     }
105     if( node[x].deep > node[y].deep ) swap(x,y);
106     if(x != y)
107         ans = max(ans,query(1,n,1,node[x].pos+1,node[y].pos));
108     return ans;
109 }
110
111
112 void Solve()
113 {
114     char ch[10];
115     int a , b; T[1].val = -inf;T[1] = (seg){0,0,-inf};
116     for(int i = 1; i < n ; ++i)
117     {
118         if(node[E[i].fro].deep > node[E[i].to].deep) swap(E[i].fro,E[i].to);
119         int now = 1;
120         updata(1,n,now,node[E[i].to].pos,E[i].val);
121     }
122     while(~scanf("%s",ch)&&ch[0]!=‘D‘)
123     {
124         if(ch[0] == ‘Q‘)
125         {
126             scanf("%d%d",&a,&b);
127             printf("%d\n",lca(a,b));
128         }
129         else if(ch[0] == ‘C‘)
130         {
131             scanf("%d%d",&a,&b);
132             int now = 1 ;
133             updata(1,n,now,node[E[a].to].pos,b);
134         }
135
136     }
137 }
138
139 int main( )
140 {
141     int t;
142     scanf("%d",&t);
143     while(t--)
144     {
145         Init();
146         Solve();
147     }
148     return 0;
149 }