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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#814073#9881. Diverge and Convergeucup-team3161#RE 1ms4008kbC++171.9kb2024-12-14 15:00:062024-12-14 15:00:08

Judging History

你现在查看的是最新测评结果

  • [2024-12-14 15:00:08]
  • 评测
  • 测评结果:RE
  • 用时:1ms
  • 内存:4008kb
  • [2024-12-14 15:00:06]
  • 提交

answer

#include<bits/stdc++.h>
#define For(i,x,y) for (int i=(x);i<=(y);i++)
#define FOR(i,x,y) for (int i=(x);i<(y);i++)
#define Dow(i,x,y) for (int i=(x);i>=(y);i--)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<int,int> pa;

inline ll read(){
    ll x;
    scanf("%lld",&x);
    return x;
}

const int N = 1010;
int n,ans2[N];
pa e[N],ee[N];
set<pa>e1[N],e2[N];

int col[N];
inline void dfs(int u,int fa,int c){
    col[u]=c;
    for (auto [v,id]:e1[u]) if (v!=fa){
        dfs(v,u,c);
    }
}

int fa[N],fid[N];
inline void Dfs(int u){
    for (auto [v,id]:e2[u]) if (v!=fa[u]){
        fa[v]=u,fid[v]=id;
        Dfs(v);
    }
}

int main(){
    n=read();
    For(i,1,n-1){
        int x=read(),y=read();
        e1[x].insert(mp(y,i));
        e1[y].insert(mp(x,i));
        e[i]=mp(x,y);
    }
    For(i,1,n-1){
        int x=read(),y=read();
        e2[x].insert(mp(y,i));
        e2[y].insert(mp(x,i));
        ee[i]=mp(x,y);
    }
    For(i,1,n-1){
        auto [x,y]=e[i];
        dfs(x,y,0);
        dfs(y,x,1);
        bool flag=0;
        fa[x]=0,fid[x]=0,Dfs(x);
        for (int u=y;u;u=fa[u]){
            int v=fa[u];
            if (v){
                if (col[u]!=col[v]&&fid[u]>0){
                    flag=1;
                    ans2[i]=fid[u];
                    break;
                }
            }
        }
        assert(flag);
        e1[x].erase(mp(y,i));
        e1[y].erase(mp(x,i));
        e2[x].insert(mp(y,-i));
        e2[y].insert(mp(x,-i));
        auto [x2,y2]=ee[ans2[i]];
        e2[x2].erase(mp(y2,ans2[i]));
        e2[y2].erase(mp(x2,ans2[i]));
        e1[x2].insert(mp(y2,-ans2[i]));
        e1[y2].insert(mp(x2,-ans2[i]));
        //printf("%d %d\n",i,ans2[i]);
    }
    For(i,1,n-1) printf("%d ",i); 
    puts("");
    For(i,1,n-1) printf("%d ",ans2[i]);
    puts("");
}

Details

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Test #1:

score: 100
Accepted
time: 1ms
memory: 3920kb

input:

4
1 2
2 3
3 4
1 2
2 4
2 3

output:

1 2 3 
1 3 2 

result:

ok Correct, N = 4

Test #2:

score: 0
Accepted
time: 0ms
memory: 4008kb

input:

2
1 2
2 1

output:

1 
1 

result:

ok Correct, N = 2

Test #3:

score: 0
Accepted
time: 0ms
memory: 3996kb

input:

7
1 2
1 3
2 4
2 5
3 6
3 7
1 5
1 6
1 7
5 2
6 3
7 4

output:

1 2 3 4 5 6 
1 2 6 4 5 3 

result:

ok Correct, N = 7

Test #4:

score: -100
Runtime Error

input:

1000
780 67
364 281
934 245
121 472
460 233
574 534
91 687
91 832
413 613
815 638
519 196
992 120
150 157
198 365
643 700
279 698
623 215
578 330
869 333
874 570
879 697
897 671
962 670
108 137
779 9
988 91
919 314
696 722
431 270
810 692
769 49
254 915
737 580
229 888
379 612
19 161
787 346
280 753...

output:


result: