2015 Multi-University Training Contest 1 - 1009 Annoying problem

 Annoying problem#

Problem's Link:  http://acm.hdu.edu.cn/showproblem.php?pid=5296  
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Mean: 

给你一个有根树和一个节点集合S，初始S为空，有q个操作，每个操作要么从树中选一个结点加入到S中（不删除树中节点），要么从S集合中删除一个结点。你需要从树中选一些边组成集合E，E中的边能够是S中的节点连通。对于每一个操作，输出操作后E中所有边的边权之和。

analyse:

首先是构图，把树看作一个无向图，使用邻接表存图。

处理出从1号结点的dfs序存储起来。

添加点u操作：查找S集合中的点与添加的点dfs序在前面且编号最大的点，以及dfs序在后面且编号最小的点，设这两个点是x,y。

那么增加的花费是：dis[u]-dis[lca[(x,u)] - dis [lca(y,u)] + dis[lca(x,y) ]; 其中dis代表该点到根节点的距离。

 

对于删除点u操作：先把点从集合中删除，然后再计算减少花费，计算公式和增加的计算方法一样。

也是看了题解才撸出来的。

Time complexity: O(N)

 

Source code: 

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/* * this code is made by crazyacking * Verdict: Accepted * Submission Date: 2015-07-22-11.22 * Time: 0MS * Memory: 137KB */ #include <queue> #include <cstdio> #include <set> #include <string> #include <stack> #include <cmath> #include <climits> #include <map> #include <cstdlib> #include <iostream> #include <vector> #include <algorithm> #include <cstring> #define  LL long long #define  ULL unsigned long long #define rep(i,n) for(int i=0;i<n;++i) using namespace std;   const int N = 100010,D=20; int st[N], ori[N], dfs_clock; vector<pair<int, int> > G[N]; int shortcut[D][N], dep[N]; int *fa; int get_lca( int a, int b ) {       if( dep[a] < dep[b] )             swap( a, b );       for( int i = D - 1; ~i; --i )       {             if( dep[a] - dep[b] >> i & 1 )                   a = shortcut[i][a];       }       if( a == b ) return a;       for( int i = D - 1; ~i; --i )       {             if( shortcut[i][a] != shortcut[i][b] )             {                   a = shortcut[i][a];                   b = shortcut[i][b];             }       }       return fa[a]; }   int dis[N]; void dfs( int u, int father ) {       st[u] = ++dfs_clock;       ori[dfs_clock] = u;       vector<pair<int, int> > :: iterator it;       for( it = G[u].begin(); it != G[u].end(); ++it )       {             int v = ( *it ).first;             int w = ( *it ).second;             if( v == father )continue;             fa[v] = u;             dep[v] = dep[u] + 1;             dis[v] = dis[u] + w;             dfs( v, u );       } }     void prepare( int n ) {       for( int j = 1; j < D; ++j )       {             rep( i, n )             {                   int &res = shortcut[j][i];                   res = shortcut[j - 1][i];                   if( ~res ) res = shortcut[j - 1][res];             }       } }     set<int> nodes; int get_dis( int a, int b ) {       return dis[a] + dis[b] - 2 * dis[get_lca( a, b )]; }     int add( int u ) {       if( !nodes.empty() )       {             set<int>::iterator low, high;             low = nodes.lower_bound( st[u] );             high = low;             if( low == nodes.end() || low == nodes.begin() )             {                   low = nodes.begin();                   high = nodes.end();                   high--;             }             else low--;             int x = ori[*low];             int y = ori[*high];             int res = get_dis( x, u ) + get_dis( y, u ) - get_dis( x, y );             return res;       }       return 0; }   int main() {       ios_base::sync_with_stdio( false );       cin.tie( 0 );       int T, n, q, u, v, w;       scanf( "%d", &T );       rep( cas, T )       {             scanf( "%d %d", &n, &q );             rep( i, n ) G[i].clear();             dfs_clock = 0;             rep( i, n - 1 )             {                   scanf( "%d %d %d", &u, &v, &w );                   u--;                   v--;                   G[u].push_back( make_pair( v, w ) );                   G[v].push_back( make_pair( u, w ) );             }             fa = shortcut[0];             fa[0] = -1;             dfs( 0, -1 );             prepare( n );             nodes.clear();             printf( "Case #%d:\n", cas + 1 );             int ans = 0;             while( q-- )             {                   int op;                   scanf( "%d %d", &op, &u );                   u--;                   if( op == 1 )  // add                   {                         if( nodes.find( st[u] ) == nodes.end() )                         {                               ans += add( u );                               nodes.insert( st[u] );                         }                   }                   else   // delete                   {                         if( nodes.find( st[u] ) != nodes.end() )                         {                               nodes.erase( st[u] );                               ans -= add( u );                         }                   }                   printf( "%d\n", ans >> 1 );             }       }       return 0; } /*   */