• spoj375 QTREE

    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
",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 }
  • 相关阅读:
    Vue 使用Scss，深度修改局部样式
    Sublime Text 插件：批量删除空白行
    Sublime Text 3前端开发常用优秀插件介绍
    常用的sublime text 3插件（很爽哦）
    20 个强大的 Sublime Text 插件
    Java多线程之synchronized详解
    进阶Java多线程
    初识Java多线程
    分布式锁实现的正确打开方式
    分布式session实现方式
  • 原文地址：https://www.cnblogs.com/Ateisti/p/6308586.html
Copyright © 2020-2023  润新知