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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#143111 | #6502. Disjoint Set Union | qzez | WA | 1ms | 3748kb | C++14 | 2.1kb | 2023-08-20 16:51:41 | 2023-08-20 16:51:43 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
template<typename T>
ostream& operator << (ostream &out,const vector<T>&x){
if(x.empty())return out<<"[]";
out<<'['<<x[0];
for(int len=x.size(),i=1;i<len;i++)out<<','<<x[i];
return out<<']';
}
template<typename T>
vector<T> ary(const T *a,int l,int r){
return vector<T>{a+l,a+1+r};
}
template<typename T>
void debug(T x){
cerr<<x<<'\n';
}
template<typename T,typename ...S>
void debug(T x,S ...y){
cerr<<x<<' ',debug(y...);
}
const int N=1e3+10;
int T,n,fa[N],a[N],col[N];
int find(int x){
return fa[x]==x?x:fa[x]=find(fa[x]);
}
vector<tuple<int,int,int> >ans;
void merge(int x,int y){
ans.push_back({2,x,y});
x=find(x),y=find(y);
if(x^y)fa[x]=y;
}
int k,in[N],id[N];
vector<int>to[N];
bool topo(){
queue<int>q;
for(int i=1;i<=n;i++){
if(!in[i])q.push(i);
}
k=0;
for(int u;!q.empty();){
id[++k]=u=q.front(),q.pop();
for(int v:to[u]){
if(!--in[v])q.push(v);
}
}
return k==n;
}
bool get(){
scanf("%d",&n);
for(int i=1;i<=n;i++)scanf("%d",&fa[i]);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
if(a[i]^i)to[a[i]].push_back(i),in[i]++;
}
if(!topo())return 0;
for(int i=1;i<=n;i++){
for(col[i]=i;a[col[i]]^col[i];col[i]=a[col[i]]);
}
// debug("OK1");
// for(int i=1;i<=n;i++)debug(ary(anc[i],1,n));
debug(ary(id,1,n));
debug(ary(col,1,n));
for(int i=n,u;u=id[i],i>=1;i--){
if(fa[u]^u)continue;
for(int j=i+1,v;v=id[j],j<=n;j++){
if(col[u]==col[v]&&fa[v]==v)merge(v,u);
}
for(int v=1;v<=n;v++){
if(fa[v]!=a[v]&&fa[v]!=v&&fa[v]!=u&&fa[fa[v]]==u){
merge(v,u);
}
}
debug(u,ary(fa,1,n));
}
// debug("OK2");
for(int i=1;i<=n;i++)if(fa[i]^a[i])return 0;
puts("YES");
printf("%d\n",(int)ans.size());
for(auto x:ans){
if(get<0>(x)==1)printf("%d %d\n",get<0>(x),get<1>(x));
else printf("%d %d %d\n",get<0>(x),get<1>(x),get<2>(x));
}
return 1;
}
void clr(){
ans.clear();
for(int i=1;i<=n;i++)to[i].clear(),in[i]=0;
// for(int i=1;i<=n;i++)fill(anc[i]+1,anc[i]+1+n,0);
}
int main(){
for(scanf("%d",&T);T--;clr())if(!get())puts("NO");
return 0;
}
详细
Test #1:
score: 0
Wrong Answer
time: 1ms
memory: 3748kb
input:
5 3 1 2 3 2 2 3 4 1 2 3 3 1 1 1 2 5 1 2 3 4 5 2 3 4 5 5 5 1 1 1 1 1 1 2 3 4 5 6 1 2 2 4 5 6 1 1 5 1 4 2
output:
YES 1 2 1 2 YES 4 2 3 2 2 4 2 2 2 1 2 3 1 YES 4 2 1 2 2 2 3 2 3 4 2 4 5 NO NO
result:
wrong answer you didn't find a solution but jury did (test case 5)