SPOJ2666. Query on a tree IV
SPOJ Problem Set (classical)2666. Query on a tree IVProblem code: QTREE4 |
You are given a tree (an acyclic undirected connected graph) with N nodes, and nodes numbered 1,2,3...,N. Each edge has an integer value assigned to it(note that the value can be negative). Each node has a color, white or black. We define dist(a, b) as the sum of the value of the edges on the path from node a to node b.
All the nodes are white initially.
We will ask you to perfrom some instructions of the following form:
- C a : change the color of node a.(from black to white or from white to black)
- A : ask for the maximum dist(a, b), both of node a and node b must be white(a can be equal to b). Obviously, as long as there is a white node, the result will alway be non negative.
Input
- In the first line there is an integer N (N <= 100000)
- In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of value c (-1000 <= c <= 1000)
- In the next line, there is an integer Q denotes the number of instructions (Q <= 100000)
- In the next Q lines, each line contains an instruction "C a" or "A"
Output
For each "A" operation, write one integer representing its result. If there is no white node in the tree, you should write "They have disappeared.".
Example
Input: 3 1 2 1 1 3 1 7 A C 1 A C 2 A C 3 A Output: 2 2 0 They have disappeared.
--------------------------------------------------------
题目大意：见漆子超论文。
解题思路：见漆子超论文。吐槽吐槽，我写了两天的代码，都怪自己代码能力太弱，看懂了论文，知道了过程，无奈写不出来。
/*
题目来源：SPOJ2666. Query on a tree IV
题目难度：难题+神题
解题思路：树链剖分+线段树维护
漆子超论文里面的原题，思路完全从那里看过来的，无奈树链剖分以及线段树的维护太过于
繁琐，编码复杂，写了两天。因为这道题的树链剖分实际上是应用了树的链的分治，对于每
条重链都要建立一颗线段树，那么重链的个数在极限情况下有O(n)个，都不可能对每个线段
树开单独的数组来维护，考虑到所有线段树的节点之和其实不会超过6*N个，我就开了一个
储存线段树的数组，对于每个不同的线段树，我设置一个漂移量pn[t]，那么t这个线段树的
根节点在数组中的实际位置是pn[t]+1，另外还有很多奇怪的序号需要维护：son[i]代表i
的重链儿子所在的边，fa[i]代表i节点的父亲节点所在的边，top[i]表示i所在的重链的链头，
siz[i]表示i的子孙节点的数量，cp[i]表示i所在的重链的编号，cd[i]表示i在自己重链上
的第几个位置，csz[i]表示编号为i的重链的节点数量，pn[i]表示编号为i的重链的漂移量，
cid[i]表示它所在的重链t上第(i-pn[t])上的点在树中的编号，cnum重链计数，pnum漂移
量计数。
代码极其丑陋，可以排进我的十大最丑陋代码了
*/
#include <stdio.h>
#include <string.h>
#include <queue>
#define clr(a,b) memset(a,b,sizeof(a))
using namespace std;
const int N=200005,M=7*N,inf=0x7fffffff;
struct Node{
int v,p;
bool operator<(const Node & a)const{
return v<a.v;
}
Node(int a,int b):v(a),p(b){}
};
int n,eid;
int head[N],ed[N<<1],val[N<<1],nxt[N<<1];
int son[N],fa[N],top[N],siz[N],dis[N],col[N];
int cp[N],cd[N],csz[N],pn[N],cid[M],cnum,pnum;
int le[M],ri[M],maxl[M],maxr[M],opt[M];
priority_queue<Node>que[N],allque;
void addedge(int s,int e,int v){
ed[eid]=e;val[eid]=v;nxt[eid]=head[s];head[s]=eid++;
}
int pl(int a,int b){
if(a==-inf||b==-inf)return -inf;
return a+b;
}
void dfs_1(int s,int f){
siz[s]=1;int maxx=0,p=-1;top[s]=s;fa[s]=f;
for(int i=head[s];~i;i=nxt[i]){
int e=ed[i];if(e==f){fa[s]=i;continue;}
dfs_1(e,s);siz[s]+=siz[e];
if(siz[e]>maxx){maxx=siz[e];p=i;}
}
son[s]=p;
}
void buildtree(int p,int rt,int l,int r){
le[p+rt]=l;ri[p+rt]=r;int mid=(l+r)>>1;
maxl[p+rt]=maxr[p+rt]=opt[p+rt]=-inf;
if(l==r)return ;
buildtree(p,rt<<1,l,mid);
buildtree(p,rt<<1|1,mid+1,r);
}
void update(int p,int rt,int x){
int l=le[p+rt],r=ri[p+rt],mid=(l+r)>>1;
int root=p+rt,lroot=p+(rt<<1),rroot=p+(rt<<1|1);
if(l==x&&r==x){
int s=cid[p+x],d1=-inf,d2=-inf,p1=-1;
while(!que[s].empty()){
int v=que[s].top().v,ps=que[s].top().p;que[s].pop();
int e=cid[ps];
if(maxl[ps]+val[fa[e]]!=v)continue;
p1=ps;d1=v;break;
}
while(!que[s].empty()){
int v=que[s].top().v,ps=que[s].top().p,e=cid[ps];
if(maxl[ps]+val[fa[e]]!=v||ps==p1){que[s].pop();continue;}
d2=v;break;
}
if(~p1)que[s].push(Node(d1,p1));
if(col[s])maxl[root]=maxr[root]=max(d1,0);
else maxl[root]=maxr[root]=d1;
if(col[s])opt[root]=max(d1,pl(d1,d2));
else opt[root]=pl(d1,d2);
if(col[s])opt[root]=max(opt[root],0);
if(rt==1){
if(~fa[s]){
int f=ed[fa[s]],v=val[fa[s]];
if(maxl[root]!=-inf)que[f].push(Node(maxl[root]+v,root));
}
if(opt[root]!=-inf)allque.push(Node(opt[root],root));
}
return ;
}
if(x<=mid)update(p,rt<<1,x);
else update(p,rt<<1|1,x);
int dm0=dis[cid[p+mid]],dm1=dis[cid[p+mid+1]];
int dr=dis[cid[p+r]],dl=dis[cid[p+l]];
maxl[root]=max(maxl[lroot],pl(dm1-dl,maxl[rroot]));
maxr[root]=max(maxr[rroot],pl(dr-dm0,maxr[lroot]));
opt[root]=max(opt[lroot],opt[rroot]);
opt[root]=max(opt[root],pl(pl(maxl[rroot],maxr[lroot]),dm1-dm0));
int s=cid[p+1];
if(rt==1){
if(~fa[s]){
int f=ed[fa[s]],v=val[fa[s]];
if(maxl[root]!=-inf)que[f].push(Node(maxl[root]+v,root));
}
if(opt[root]!=-inf)allque.push(Node(opt[root],root));
}
}
void dfs_2(int s,int d,int c){
if(s==top[s]){cp[s]=++cnum;csz[cp[s]]=0;}
if(~son[s]){
top[ed[son[s]]]=top[s];dfs_2(ed[son[s]],d+val[son[s]],c+1);
}
for(int i=head[s];~i;i=nxt[i])
if(i!=son[s]&&i!=fa[s])dfs_2(ed[i],0,1);
cp[s]=cp[top[s]];cd[s]=c;int k=cp[s];
csz[k]=max(csz[k],cd[s]);dis[s]=d;
if(s==top[s]){
pn[k]=pnum;pnum+=6*csz[k];
int t=s,num=2;cid[pn[k]+1]=s;
while(~son[t]){
cid[pn[k]+num]=ed[son[t]];
t=ed[son[t]];num++;
}
}
}
void dfsinit(int s){
col[s]=1;
if(~son[s])dfsinit(ed[son[s]]);
for(int i=head[s];~i;i=nxt[i])
if(i!=fa[s]&&i!=son[s])dfsinit(ed[i]);
update(pn[cp[s]],1,cd[s]);
}
int main(){
// freopen("/home/axorb/in","r",stdin);
// freopen("/home/axorb/out","w",stdout);
scanf("%d",&n);eid=0;clr(head,-1);
for(int i=1;i<n;i++){
int a,b,c;scanf("%d%d%d",&a,&b,&c);
addedge(a,b,c);addedge(b,a,c);
}
dfs_1(1,-1);cnum=pnum=0;dfs_2(1,0,1);clr(col,0);
while(!allque.empty())allque.pop();
for(int i=1;i<=cnum;i++){
while(!que[i].empty())que[i].pop();buildtree(pn[i],1,1,csz[i]);
}
dfsinit(1);
// printf("%d\n",allque.top().v);
int q;scanf("%d",&q);
while(q--){
char ss[20];scanf("%s",ss);
if(ss[0]=='A'){
int flag=0;
while(!allque.empty()){
int v=allque.top().v,p=allque.top().p;
if(opt[p]!=v){allque.pop();continue;}
printf("%d\n",v);flag=1;break;
}
if(!flag)puts("They have disappeared.");
}
else{
int x;scanf("%d",&x);col[x]^=1;
while(top[x]!=1){
update(pn[cp[x]],1,cd[x]);x=ed[fa[top[x]]];
}
update(pn[cp[x]],1,cd[x]);
}
}
}