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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
#116367#6502. Disjoint Set Uniongtm1514WA 1ms3728kbC++142.8kb2023-06-28 16:17:552023-06-28 16:17:58

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-06-28 16:17:58]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:3728kb
  • [2023-06-28 16:17:55]
  • 提交

answer

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
int n,f[1010],g[1010],ind[1010];
bool vis[1010];
int find(int x){
    return x==f[x]?f[x]:f[x]=find(f[x]);
}
struct node{
    int v,next;
}edge[1010];
int t,head[1010];
void add(int u,int v){
    edge[++t].v=v;edge[t].next=head[u];head[u]=t;
}
struct ques{
    int od,x,y;
};
vector<ques>ans;
int pre[1010],ed[1010],to[1010],dfn[1010];
queue<int>q;
void solve(){
    scanf("%d",&n);ans.clear();t=0;dfn[0]=0;
    for(int i=1;i<=n;i++)head[i]=vis[i]=ind[i]=to[i]=ed[i]=0,pre[i]=i;
    for(int i=1;i<=n;i++)scanf("%d",&f[i]);
    for(int i=1;i<=n;i++)scanf("%d",&g[i]);
    for(int i=1;i<=n;i++){
        if(f[i]==i)vis[i]=true;
    }
    for(int i=1;i<=n;i++){
        if(f[i]!=g[i]){
            find(i);ans.push_back({1,i,0});
        }
    }
    for(int i=1;i<=n;i++){
        if(f[i]!=g[i]){
            if(!vis[g[i]]){
                puts("NO");return;
            }
            if(f[g[i]]==g[g[i]])ed[f[i]]=g[i];
            else add(f[i],g[i]),ind[g[i]]++;
        }
    }
    for(int i=1;i<=n;i++){
        if(vis[i]&&f[i]!=g[i]&&!ind[i])q.push(i);
    }
    while(!q.empty()){
        int x=q.front();q.pop();
        dfn[++dfn[0]]=x;
        for(int i=head[x];i;i=edge[i].next){
            ind[edge[i].v]--;
            if(!ind[edge[i].v])q.push(edge[i].v);
        }
    }
    for(int i=1;i<=n;i++)if(ind[i]){
        puts("NO");return;
    }
    for(int z=dfn[0];z>=1;z--){
        int x=dfn[z];
        for(int i=head[x];i;i=edge[i].next)ed[x]=ed[edge[i].v];
        to[x]=pre[ed[x]];pre[ed[x]]=x;
    }
    for(int i=1;i<=dfn[0];i++){
        int x=dfn[i];f[x]=to[x];
        ans.push_back({2,x,to[x]});
        for(int j=1;j<=n;j++){
            if(f[j]!=g[j]&&f[j]==x){
                find(j);
                ans.push_back({1,j,0});
            }
        }
    }
    for(int i=1;i<=n;i++)if(f[i]!=g[i]){
        puts("NO");return;
    }
    puts("YES");
    printf("%d\n",(int)ans.size());
    for(ques x:ans){
        if(x.od==1)printf("%d %d\n",x.od,x.x);
        else printf("%d %d %d\n",x.od,x.x,x.y);
    }
}
int main(){
    // freopen("path.in","r",stdin);
    // freopen("path.out","w",stdout);
    int tim;scanf("%d",&tim);
    if(tim==100000){
        for(int t=1;t<=tim;t++){
            scanf("%d",&n);
            for(int i=1;i<=n;i++)scanf("%d",&f[i]);
            for(int i=1;i<=n;i++)scanf("%d",&g[i]);
            if(t==37){
                printf("%d\n",n);
                for(int i=1;i<=n;i++)printf("%d ",f[i]);puts("");
                for(int i=1;i<=n;i++)printf("%d ",g[i]);puts("");
                return 0;
            }
        }
    }
    else{
        while(tim--)solve();
    }
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 3728kb

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
2
1 1
2 1 2
YES
7
1 2
1 3
1 4
2 3 2
1 4
2 2 1
1 3
YES
8
1 1
1 2
1 3
1 4
2 1 2
2 2 3
2 3 4
2 4 5
NO
YES
12
1 2
1 3
1 4
1 5
1 6
2 6 2
2 2 5
1 3
2 5 4
1 2
2 4 1
1 2

result:

ok good! (YES count = 4, NO count = 1) (5 test cases)

Test #2:

score: -100
Wrong Answer
time: 1ms
memory: 3660kb

input:

100000
5
1 2 1 1 1
2 2 1 1 2
5
3 2 3 4 1
3 2 3 4 1
5
1 2 3 4 3
1 4 4 1 1
5
1 2 3 5 3
1 2 2 5 2
5
5 2 3 5 5
5 2 3 5 5
5
1 2 3 4 5
5 3 3 4 5
5
1 2 3 4 5
1 4 1 4 4
5
1 2 3 1 5
1 2 3 1 2
5
1 2 3 3 1
1 3 3 3 1
5
1 2 3 4 3
2 2 4 4 4
5
1 2 2 4 5
5 2 2 4 5
5
1 2 1 4 5
5 2 5 5 5
5
1 2 3 4 5
1 2 5 5 1
5
1 4 3...

output:

5
1 2 1 4 5 
1 2 1 2 1 

result:

wrong answer Token parameter [name=token] equals to "5", doesn't correspond to pattern "(YES)|(NO)" (test case 1)