2015 Multi-University Training Contest 1
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Annoying problem
Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=5296
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:
/** 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;}/**/
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