鏈接:https://vjudge.net/problem/ZOJ-3261
思路:有幾天沒(méi)做并查集了吐葱,這個(gè)題我們先按權(quán)值合并街望,小的向大的合并,如果權(quán)值相等我們考慮編號(hào)弟跑,我們將所有命令先讀入灾前,然后在從末態(tài)開(kāi)始輸出詢問(wèn),destroy就變成了建邊孟辑,每次合并時(shí)同上合并方式哎甲,詢問(wèn)時(shí)檢查一下當(dāng)前點(diǎn)的權(quán)值和父節(jié)點(diǎn)權(quán)值是否相等,如果相等則無(wú)解饲嗽,否則記錄答案(一定要先getroot一下啊不然上次更新的并沒(méi)有傳遞上去炭玫,這個(gè)地方已經(jīng)錯(cuò)了很多次了長(zhǎng)點(diǎn)記性啊拜托!C蚕骸M碳印!)
代碼:
#pragma GCC diagnostic error "-std=c++11"
#include<bits/stdc++.h>
using namespace std;
const int maxn = 60010;
int n,m,q;
set<pair<int,int> > edge;
struct order{
int o,u,v;
}ss[maxn];
int maxv[maxn];
int par[maxn];
int ans[maxn];
char ch[100];
int getroot(int a){
if(par[a]==a)return a;
return par[a] = getroot(par[a]);
}
void merge(int u,int v){//先按權(quán)值合并尽狠,權(quán)值相同按編號(hào)合并
int p1 = getroot(u);
int p2 = getroot(v);
if(p1==p2)return ;
if(maxv[p1]>maxv[p2]){
par[p2] = p1;
}
else if(maxv[p1]<maxv[p2]){
par[p1] = p2;
}
else{
if(p1<p2){
par[p2] = p1;
}
else if(p1>p2){
par[p1] = p2;
}
}
}
int main(){
int kase = 0;
while(~scanf("%d",&n)){
if(kase++)printf("\n");
edge.clear();
for(int i=0;i<=n;i++)par[i] = i;
for(int i=0;i<n;i++)scanf("%d",&maxv[i]);
scanf("%d",&m);
for(int i=0;i<m;i++){
int u,v;
scanf("%d%d",&u,&v);
if(u>v)swap(u,v);
edge.insert(make_pair(u,v));
}
scanf("%d",&q);
for(int i=0;i<q;i++){
int a,b;
scanf("%s%d",ch,&a);
if(ch[0]=='q'){
ss[i].o = 0;
ss[i].u = a;
}
else if(ch[0]=='d'){
scanf("%d",&b);
if(a>b)swap(a,b);
ss[i].o = 1;
ss[i].u = a;
ss[i].v = b;
edge.erase(make_pair(a,b));
}
}
for(auto &it:edge){//將現(xiàn)村的邊進(jìn)行更新
int u = it.first;
int v = it.second;
merge(u,v);
}
for(int i=q-1;i>=0;i--){
if(!ss[i].o){
getroot(ss[i].u);//O魏!0栏唷<肌!一定要記得先路徑壓縮一下更新最新的父節(jié)點(diǎn)3凉荨B氲场!:芳啊闽瓢!
if(maxv[ss[i].u]!=maxv[par[ss[i].u]])
ans[i] = par[ss[i].u];
else ans[i] = -1;
}
else{
ans[i] = -2;
int u = ss[i].u;
int v = ss[i].v;
merge(u,v);
}
}
for(int i=0;i<q;i++){
if(ans[i]!=-2)
printf("%d\n",ans[i]);
}
}
return 0;
}