【codeforces】487E. Tourists 点双连通+树链剖分

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题目分析：这么多天了终于艹掉了这题，太感动了T ^ T

看到这题，询问路径最小值问题，第一反应就是缩成一棵树然后树链剖分求解。兴冲冲的敲了一个边双连通，测了样例没过，发现不可以用边双连通。。。然后换成点双连通，将一个点双连通块中的边缩成一个点，然后所有的割顶缩成一个点，然后一个连通块我们用一个set保存最小值。在没有更新的情况下，就是不断查询树上路径的最小值。当然如果查询的两个点是同一个点时，只能输出这个点的值，因为题目要求路径上的点不能重复。

然后，当涉及到修改的时候，又被卡住了，由于一个割顶的修改会影响到和他直接相连的所有连通块（假设图为星形），这样修改所有与其相连的复杂度太高了！。

有什么方法可以保证复杂度呢？

很幸运，方法是有的！当每次修改时，修改信息的为割顶时，我们就将割顶的父亲（一个点连通块）上的信息同时修改了！当每次查询时，如果查询路径的lca是点连通块，则同时查询该点连通块的父亲（割顶），这样我们的算法复杂度就得以保证了。

PS：注意tarjan求点连通块缩边后一定是割顶所连的一定是点连通块，点连通块所连的一定是割顶。

做的时候有段时间一直WA5。。。本地跑出来的和cf的不同也是很蛋疼，虽然知道一定是算法错了，可是就是加了特判，竟然跑的时候特判被忽视了！MLE13的时候发现树上有许多的重边。。导致树额外多了N个莫名其妙节点。。

代码如下：

#include <set>#include <vector>#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;typedef long long LL ;#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )#define clr( a , x ) memset ( a , x , sizeof a )#define ls ( o << 1 )#define rs ( o << 1 | 1 )#define lson ls , l , m#define rson rs , m + 1 , r#define root 1 , 1 , G.bcc_cnt#define mid ( ( l + r ) >> 1 )const int MAXN = 200005 ;const int MAXE = 400005 ;const int INF = 0x3f3f3f3f ;struct Edge {int v , f , n ;Edge () {}Edge ( int v , int f , int n ) : v ( v ) , f ( f ) , n ( n ) {}} ;struct BCC {Edge E[MAXE] ;int H[MAXN] , cntE ;int dfn[MAXN] , low[MAXN] , dfs_clock ;int bcc[MAXN] , bcc_cnt ;int is[MAXN] ;int block[MAXN] ;int ncnt ;int S[MAXN] , top ;void clear () {top = 0 ;cntE = 0 ;bcc_cnt = 0 ;dfs_clock = 0 ;clr ( H , -1 ) ;clr ( dfn , 0 ) ;clr ( is , 0 ) ;}void addedge ( int u , int v ) {E[cntE] = Edge ( v , 0 , H[u] ) ;H[u] = cntE ++ ;}void tarjan ( int u , int fa = 0 ) {dfn[u] = low[u] = ++ dfs_clock ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( E[i].f ) continue ;E[i].f = E[i ^ 1].f = 1 ;S[top ++] = ( i >> 1 ) ;if ( !dfn[v] ) {tarjan ( v , u ) ;low[u] = min ( low[v] , low[u] ) ;if ( low[v] >= dfn[u] ) {++ is[u] ;++ bcc_cnt ;while ( 1 ) {int x = S[-- top] ;bcc[x] = bcc_cnt ;if ( x == ( i >> 1 ) ) break ;}}} else low[u] = min ( low[u] , dfn[v] ) ;}if ( fa == 0 ) -- is[u] ;}void find_bcc ( int n ) {For ( i , 1 , n ) if ( !dfn[i] ) tarjan ( i ) ;ncnt = bcc_cnt ;For ( i , 1 , n ) if ( is[i] ) block[i] = ++ bcc_cnt ;For ( i , 1 , n ) {int u = i ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( !is[u] ) block[u] = bcc[( i >> 1 )] ;if ( !is[v] ) block[v] = bcc[( i >> 1 )] ;}}}} ;struct Heavy_Light_Decompose {Edge E[MAXE] ;int H[MAXN] , cntE ;int pos[MAXN] ;int pre[MAXN] ;int top[MAXN] ;int son[MAXN] ;int dep[MAXN] ;int siz[MAXN] ;int idx[MAXN] ;int tree_idx ;void clear () {cntE = 0 ;tree_idx = 0 ;clr ( H , -1 ) ;siz[0] = 0 ;dep[1] = 0 ;pre[1] = 0 ;}void addedge ( int u , int v ) {E[cntE] = Edge ( v , 0 , H[u] ) ;H[u] = cntE ++ ;}void dfs ( int u ) {siz[u] = 1 ;son[u] = 0 ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( v == pre[u] ) continue ;pre[v] = u ;dep[v] = dep[u] + 1 ;dfs ( v ) ;siz[u] += siz[v] ;if ( siz[v] > siz[son[u]] ) son[u] = v ;}}void rebuild ( int u , int top_element ) {top[u] = top_element ;pos[u] = ++ tree_idx ;idx[tree_idx] = u ;if ( son[u] ) rebuild ( son[u] , top_element ) ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( v != pre[u] && v != son[u] ) {rebuild ( v , v ) ;}}}} ;BCC G ;Heavy_Light_Decompose T ;set < int > st[MAXN] ;int n , m , q ;int val[MAXN] ;int vis[MAXN] ;int minv[MAXN << 2] ;inline void push_up ( int o ) {minv[o] = min ( minv[ls] , minv[rs] ) ;}void build ( int o , int l , int r ) {if ( l == r ) {minv[o] = *( st[T.idx[l]].begin () ) ;return ;}int m = mid ;build ( lson ) ;build ( rson ) ;push_up ( o ) ;}void update ( int x , int v , int o , int l , int r ) {if ( l == r ) {minv[o] = v ;return ;}int m = mid ;if ( x <= m ) update ( x , v , lson ) ;else update ( x , v , rson ) ;push_up ( o ) ;}int query ( int L , int R , int o , int l , int r ) {if ( L <= l && r <= R ) {return minv[o] ;}int m = mid ;if ( R <= m ) return query ( L , R , lson ) ;if ( m <  L ) return query ( L , R , rson ) ;return min ( query ( L , R , lson ) , query ( L , R , rson ) ) ;}int Query ( int x , int y ) {int ans = INF ;//printf ( "%d %d\n" , x , y ) ;while ( T.top[x] != T.top[y] ) {if ( T.dep[T.top[x]] < T.dep[T.top[y]] ) swap ( x , y ) ;//printf ( "--%d %d\n" , T.pos[T.top[x]] , T.pos[x] ) ;ans = min ( ans , query ( T.pos[T.top[x]] , T.pos[x] , root ) ) ;x = T.pre[T.top[x]] ;}if ( T.dep[x] > T.dep[y] ) swap ( x , y ) ;ans = min ( ans , query ( T.pos[x] , T.pos[y] , root ) ) ;//printf ( "%d\n" , ans ) ;if ( x <= G.ncnt && T.pre[x] > 0 ) {//printf ( "%233\n" ) ;ans = min ( ans , *( st[T.pre[x]].begin () ) ) ;}return ans ;}void solve () {char op[5] ;int u , v ;G.clear () ;T.clear () ;For ( i , 1 , n ) scanf ( "%d" , &val[i] ) ;rep ( i , 0 , m ) {scanf ( "%d%d" , &u , &v ) ;G.addedge ( u , v ) ;G.addedge ( v , u ) ;}G.find_bcc ( n ) ;//printf ( "%d %d\n" , G.bcc_cnt , G.ncnt ) ;clr ( vis , 0 ) ;For ( u , 1 , n ) {for ( int i = G.H[u] ; ~i ; i = G.E[i].n ) {if ( G.block[u] != G.bcc[i >> 1] && !vis[G.bcc[i >> 1]] ) {//printf ( "%d->%d\n" , G.block[u] , G.bcc[i >> 1] ) ;T.addedge ( G.block[u] , G.bcc[i >> 1] ) ;T.addedge ( G.bcc[i >> 1] , G.block[u] ) ;vis[G.bcc[i >> 1]] = 1 ;}}for ( int i = G.H[u] ; ~i ; i = G.E[i].n ) vis[G.bcc[i >> 1]] = 0 ;}T.dfs ( 1 ) ;T.rebuild ( 1 , 1 ) ;//printf ( "---_____%d\n" , T.tree_idx ) ;For ( i , 1 , G.bcc_cnt ) st[i].clear () ;For ( i , 1 , n ) {st[G.block[i]].insert ( val[i] ) ;if ( G.block[i] > G.ncnt && T.pre[G.block[i]] > 0 ) {st[T.pre[G.block[i]]].insert ( val[i] ) ;}}build ( root ) ;//printf ( "%d\n" , T.pre[G.block[2]] ) ;//printf ( "ok\n" ) ;//printf ( "%d %d\n" , T.pre[2] , T.pre[6] ) ;while ( q -- ) {scanf ( "%s%d%d" , op , &u , &v ) ;if ( op[0] == 'C' ) {//printf ( "ok\n" ) ;st[G.block[u]].erase ( val[u] ) ;st[G.block[u]].insert ( v ) ;if ( G.block[u] > G.ncnt && T.pre[G.block[u]] > 0 ) {st[T.pre[G.block[u]]].erase ( val[u] ) ;st[T.pre[G.block[u]]].insert ( v ) ;int x = *( st[T.pre[G.block[u]]].begin () ) ;update ( T.pos[T.pre[G.block[u]]] , x , root ) ;}val[u] = v ;int x = *( st[G.block[u]].begin () ) ;//printf ( "%d %d %d\n" , G.block[u] , x , val[G.blocku] ) ;update ( T.pos[G.block[u]] , x , root ) ;} else {if ( u == v ) printf ( "%d\n" , val[u] ) ;else printf ( "%d\n" , Query ( G.block[u] , G.block[v] ) ) ;}}}int main () {while ( ~scanf ( "%d%d%d" , &n , &m , &q ) ) solve () ;return 0 ;}