spoj375 QTREE - Query on a tree

  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 }