标签:des style blog color os io strong for ar
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 3836 | Accepted: 1088 |
Description
You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N − 1. Each edge is associated with a weight. Then you are to execute a series of instructions on the tree. The instructions can be one of the following forms:
CHANGE i v |
Change the weight of the ith edge to v |
NEGATE a b |
Negate the weight of every edge on the path from a to b |
QUERY a b |
Find the maximum weight of edges on the path from a to b |
Input
The input contains multiple test cases. The first line of input contains an integer t (t ≤ 20), the number of test cases. Then follow the test cases.
Each test case is preceded by an empty line. The first nonempty line of its contains N (N ≤ 10,000). The next N − 1 lines each contains three integers a, b and c, describing an edge connecting nodes a and b with weight c. The edges are numbered in the order they appear in the input. Below them are the instructions, each sticking to the specification above. A lines with the word “DONE
” ends the test case.
Output
For each “QUERY
” instruction, output the result on a separate line.
Sample Input
1 3 1 2 1 2 3 2 QUERY 1 2 CHANGE 1 3 QUERY 1 2 DONE
Sample Output
1 3
树链剖分~
然后注意一下 有n-1条边。。每条边要addedge 两次 然后,注意一下空间开两倍就可以了。否则RE
只要记录最小最大值。
取反的时候把最小最大交换一下 ,然后在取负就可以了
还有lazy更新那个取反
pushdown的时候要注意一下把孩子的lazy取反
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std; #define root 1,n,1 #define lr rt<<1 #define rr rt<<1|1 #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 const int inf=1e9; const int N=20010; struct node { int u,v,w; }e[N]; int n; int top[N],rnk[N],fa[N],p[N],son[N],siz[N],dep[N],pos; int eh[N],et[N],nxt[N],tot; void init() { memset(eh,-1,sizeof eh); memset(son,-1,sizeof son); pos=1; tot=0; } void addedge(int u,int v) { et[tot]=v;nxt[tot]=eh[u];eh[u]=tot++; et[tot]=u;nxt[tot]=eh[v];eh[v]=tot++; } void dfs1(int u,int father,int d) { fa[u]=father; dep[u]=d; siz[u]=1; for(int i=eh[u]; ~i ;i=nxt[i]){ int v=et[i]; if( v==fa[u] )continue; dfs1( v,u,d+1); if( son[u]==-1 || siz[v] > siz[ son[u] ] ) son[u]=v; } } void dfs2(int u,int tp) { top[u]=tp; p[u]=pos++; rnk[ p[u] ]=u; if( son[u] == -1 ) return ; dfs2( son[u],tp ); for(int i=eh[u]; ~i ;i=nxt[i] ){ int v=et[i]; if( v==fa[u] || v==son[u] )continue; dfs2(v,v); } } //----------------------------------- int d_m[N<<2],d_M[N<<2]; bool lazy[N<<2]; void build(int l,int r,int rt) { d_m[rt]=d_M[rt]=lazy[rt]=0; if(l==r){return ;} int mid=(l+r)>>1; build(lson); build(rson); } void Up(int rt) { d_m[rt]=min( d_m[lr],d_m[rr] ); d_M[rt]=max( d_M[lr],d_M[rr] ); } void Down(int l,int r,int rt) { if(l==r)return ; if( lazy[rt] ) { swap(d_m[rr],d_M[rr]); d_M[rr] = -d_M[rr]; d_m[rr] = -d_m[rr]; lazy[rr] ^= 1; swap(d_m[lr],d_M[lr]); d_M[lr] = -d_M[lr]; d_m[lr] = -d_m[lr]; lazy[lr] ^= 1; lazy[rt]=0; } } void update(int l,int r,int rt,int x,int v) { if(l==r){ d_m[rt]=d_M[rt]=v;lazy[rt]=0; return ; } Down(l,r,rt); int mid=(l+r) >> 1; if(x <= mid ) update(lson,x,v); else update(rson,x,v); Up(rt); } int query(int l,int r,int rt,int L,int R) { int res = -inf; if(L <= l && r<=R ){ return d_M[rt]; } Down(l,r,rt); int mid=(l+r)>>1; if( L <= mid )res=max( res,query(lson,L,R) ); if( R > mid )res=max( res,query(rson,L,R) ); return res; } void nega(int l,int r,int rt,int L,int R) { if( L <= l && r<= R ){ swap(d_m[rt],d_M[rt]); d_M[rt] = -d_M[rt]; d_m[rt] = -d_m[rt]; lazy[rt] ^= 1; return ; } Down(l,r,rt); int mid=(l+r) >> 1; if( L <= mid ) nega(lson,L,R); if( R > mid ) nega(rson,L,R); Up(rt); } int Q(int u,int v) { int res = -inf; int f1=top[u],f2=top[v]; while(f1 != f2){ if(dep[f1] < dep[f2]){ swap(f1,f2); swap(u,v); } res=max(res,query(root,p[f1],p[u])); u = fa[ f1 ]; f1 = top[ u ]; } if( u == v )return res; if( dep[u] > dep[v] )swap(u,v); return max( res , query( root,p[ son[u] ] ,p[v] ) ); } void NE(int u,int v) { int f1=top[u],f2=top[v]; while(f1 != f2){ if(dep[f1] < dep[f2]){ swap(f1,f2); swap(u,v); } nega(root,p[f1],p[u]); u = fa[ f1 ]; f1 = top[ u ]; } if( u == v )return ; if( dep[u] > dep[v] )swap(u,v); nega( root,p[ son[u] ] ,p[v]) ; } void run() { char op[10]; int x,y,v; init(); scanf("%d",&n); for(int i=1;i<n;++i){ scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w); addedge( e[i].u , e[i].v ); } dfs1(1,0,0); dfs2(1,1); build( root ); for(int i=1;i < n ;++i){ if( dep[e[i].u] > dep[ e[i].v] )swap(e[i].u,e[i].v); update(root,p[ e[i].v ],e[i].w); } while(scanf("%s",op)){ if(op[0] == ‘D‘)break; scanf("%d%d",&x,&y); if(op[0]==‘C‘){ update( root, p[ e[ x ].v ] , y ); } else if(op[0]==‘Q‘){ printf("%d\n",Q(x,y)); } else { NE(x,y); } } } int main() { int _; #ifdef LOCAL freopen("in.txt","r",stdin); #endif scanf("%d",&_); while(_--)run(); return 0; }
标签:des style blog color os io strong for ar
原文地址:http://www.cnblogs.com/YRETSIM/p/3933962.html