Meaning of the questions:
An edge weight containing a given tree and the tree of n nodes, a total of q times the operation, is divided into two
0 c: s to find the distance from the position c, and s c becomes
1 ab: the article becomes a weight of the side b
Right edge processing generally have two options:
1. The side of the point
All upcoming the n-1 sides of a node are increased, i.e. the right side of the right point, no new tree 2n-1 nodes in point to the right of the split tree chain
(Generally not
2. The edge attachment point
The edge weights attached on its end deeper depth (dep [v]> dep [u]), the processing x, y when a path between the other operations before, but does not process the final x, y point of the right of lca (since the upper right side of this path it is not represented in the right point):
The final step is: ans + = query (1, dfn [son [x]], dfn [y]); or ans + = query (1, dfn [x] +1, dfn [y]);
Note that when x when x == y, y is present Shen lca, does not operate directly on the return line
The inscribed one day also have not seen orz
First, wa: the problem is to deal with when it comes to above x == y no attention
Then has been tle, the problem is unknown, the segment into a tree structure or wording tle
Paste the code, it could be found a mistake in which:
//树链剖分 b #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #define fi first #define se second #define rep( i ,x ,y ) for( int i= x; i<= y ;i++ ) #define reb( i ,y ,x ) for( int i= y; i>= x ;i-- ) #define mem( a ,x ) memset( a ,x ,sizeof(a)) #define lson pos<<1 ,l ,m #define rson pos<<1|1 ,m+1 ,r using namespace std; typedef long long ll; typedef long double ld; typedef pair<int ,int> pii; typedef pair<ll ,ll> pll; typedef pair<string ,int> psi; const int inf = 0x3f3f3f3f; const int N = 100005; const int M = 1000005; struct Edge{ int to ,val ,next; }edge[N<<1]; struct E{ int u ,v ,c; }e[N<<1]; struct Tree{ int left ,right; int t ,lazy; }tree[N<<2]; int val[N] ,tr[N<<2]; int head[N]; int dep[N] ,dfn[N] ,son[N] ,fa[N]; int rnk[N] ,sz[N] ,top[N]; int cnt = 0,tot = 0 ,n ,q ,s ; void init( ){ cnt = tot =0; dep[1] = val[1] = 0; mem( head ,0 ); mem( son , 0 ); } void add_edge( int u ,int v ){ edge[++cnt].to = v; edge[cnt].next = head[u]; head[u] = cnt; } // sgement tree void push_up( int pos ){ tree[pos].t = tree[pos<<1].t + tree[pos<<1|1].t; } void build( int pos , int l ,int r ){ tree[pos].left = l; tree[pos].right= r; if( l==r ){ tree[pos].t = val[ rnk[l] ]; return; } int m = (l+r)>>1; build( lson ); build( rson ); push_up( pos ); } void updata( int pos ,int p ,int v ){ //cout<<tree[pos].left<<" "<<tree[pos].right<<endl; if( tree[pos].left == p && tree[pos].right ==p ){ tree[pos].t = v; return; } int m = (tree[pos].left+tree[pos].right)>>1; if( p<=m)updata( pos<<1 ,p ,v ); else updata( pos<<1|1 ,p ,v ); push_up( pos ); } int query( int pos ,int l ,int r ){ if( l <= tree[pos].left && tree[pos].right <= r){ return tree[pos].t; } int ans = 0; int m = ( tree[pos].left+tree[pos].right )>>1; if( m >= l )ans += query( pos<<1 ,l ,r ); if( m < r ) ans += query( pos<<1|1 ,l ,r ); return ans; } // tree - chain // dfs 1-- dep ,sz ,fa ,son void dfs1( int u ,int f ){ sz[u] = 1; for( int i = head[u]; i ;i = edge[i].next ){ int v = edge[i].to; if( v == f )continue; dep[v] = dep[u] + 1; fa[v] = u; dfs1( v ,u ); sz[u] += sz[v]; if( sz[v] >sz[ son[u] ] )son[u] = v; } } void dfs2( int u ,int t ){ dfn[u] = ++tot; top[u] = t; rnk[tot] = u; if( son[u] )dfs2( son[u] ,t ); for( int i = head[u] ;i ;i = edge[i].next ){ int v = edge[i].to; if( v!= fa[u] && v!= son[u] )dfs2(v,v); } } int LCA( int x ,int y ){ int fx = top[x] ,fy = top[y]; while( fx != fy ){ if( dep[x] < dep[y] )swap(x ,y) ,swap( fx ,fy ); x= fa[fx] ,fx = top[x]; } return dep[x] < dep[y] ? x : y ; } /*void c_updata( int x ,int y ,int v ){ int fx = top[x] ,fy = top[y]; while( fx != fy ){ if( dep[fx] < dep[fy] )swap(x ,y) ,swap(fx ,fy); updata( 1 ,1 ,n ,dfn[fx] ,dfn[x] ,v); x = f[x] ,fx = top[x]; } if( dfn[x] > dfn[y] )swap( x ,y ); updata( 1 ,1 ,n ,dfn[x] ,dfn[y] ,v); } */ int c_query( int x ,int y ){ int fx = top[x] ,fy = top[y]; int ans = 0; while( fx != fy ){ if( dep[fx] < dep[fy] )swap(x ,y) ,swap(fx ,fy); ans += query( 1 ,dfn[fx] ,dfn[x] ); x = fa[x] ,fx = top[x]; } if( x==y )return ans; if( dfn[x] > dfn[y] )swap( x ,y ); //cout<<x<<" son2 "<<son[x]<<endl; // x = son[x]; 致死错误 if(dfn[x] != dfn[y])ans += query( 1 ,dfn[son[x]] ,dfn[y] ); return ans ; } void rev_edge( ){ rep( i ,1 ,n-1 ){ if( dep[e[i].v] > dep[e[i].u] )val[e[i].v] = e[i].c; else val[e[i].u] = e[i].c; //cout<<" rev_edge "<<e[i].u<<" "<<val[e[i].u]<<" "<<e[i].v<<" "<<val[e[i].v]<<endl; } } int main( ){ scanf( "%d%d%d" ,&n ,&q ,&s ) ; //init(); rep( i ,1 ,n-1 ){ scanf("%d%d%d" ,&e[i].v ,&e[i].u ,&e[i].c); add_edge( e[i].u ,e[i].v); add_edge( e[i].v ,e[i].u); } dfs1( 1 ,1 ); dfs2( 1 ,1 ); rev_edge( ); build( 1 ,1 ,n ); int op ,l ,r ,z; while( q-- ){ scanf("%d" ,&op ); if( op ){ scanf("%d%d",&l ,&z); if( dep[e[l].v] > dep[e[l].u] )l = e[l].v; else l = e[l].u; updata( 1 ,dfn[l] ,z ); } else{ scanf("%d" ,&l ); printf( "%d\n" ,c_query( s ,l) ); s = l; } } return 0; }
Others can live code: https: //www.cnblogs.com/yaoyueduzhen/p/5311230.html
I think that is attached to the right side of the node that process we slightly different, the rest is basically the same, ah, why I had not written it?
sad, caught in the confusion and great frustration