結果

問題 No.1914 Directed by Two Sequences
ユーザー chineristAC
提出日時 2022-02-06 23:11:19
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 2,600 bytes
コンパイル時間 438 ms
コンパイル使用メモリ 82,176 KB
実行使用メモリ 114,220 KB
最終ジャッジ日時 2024-12-16 01:35:51
合計ジャッジ時間 15,677 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 10 RE * 28
権限があれば一括ダウンロードができます

ソースコード

diff # 

def solve(N,A,B,M,E):
    A = [a-1 for a in A]
    B = [b-1 for b in B]
    res = set()

    edge = [set() for v in range(N)]
    for u,v in E:
        edge[u-1].add(v-1)
        edge[v-1].add(u-1)
    
    group = [-1 for v in range(N)]
    for v in range(N):
        mex = [False for i in range(len(edge[v])+1)]
        for nv in edge[v]:
            if nv < v and group[nv] < len(mex):
                mex[group[nv]] = True
        for i in range(len(mex)):
            if not mex[i]:
                group[v] = i
                break

    n = max(group) + 1
    clique = [[] for g in range(n)]
    for v in range(N):
        clique[group[v]].append(v)

    def direct(i,j):
        if i < j:
            return A[i] < B[j]
        else:
            return B[i] < A[j]

    def hamilton_path(V):
        n = len(V)
        if n <= 1:
            return V

        A = hamilton_path(V[:n//2])
        B = hamilton_path(V[n//2:])
        res = []
        bi = 0
        for ai in range(len(A)):
            while bi<len(B) and direct(B[bi],A[ai]):
                res.append(B[bi])
                bi += 1
            res.append(A[ai])
        res += B[bi:]
        return res

    idx = [-1 for v in range(N)]
    for g in range(n):
        clique[g] = hamilton_path(clique[g])
        for i in range(len(clique[g])):
            idx[clique[g][i]] = i
        for u,v in zip(clique[g],clique[g][1:]):
            res.add((u,v))
    
    memo_to = [N for v in range(N)]
    memo_from = [-1 for v in range(N)]
    ci = [i for i in range(n)]
    ci.sort(lambda g:len(clique[g]))
    Vs = []

    for i in range(n):
        target = ci[i]
        Vs += clique[target]
        
        for v in Vs:
            for nv in clique[target]:
                if v==nv:
                    continue
                if direct(v,nv) and nv not in edge[v]:
                    memo_to[v] = min(memo_to[v],idx[nv])
                if direct(nv,v) and v not in edge[nv]:
                    memo_from[v] = max(memo_from[v],idx[nv])
            if memo_to[v]!=N:
                nv = clique[target][memo_to[v]]
                res.add((v,nv))      
                memo_to[v] = N
            if memo_from[v]!=-1:
                pv = clique[target][memo_from[v]]
                res.add((pv,v))
                memo_from[v] = -1
    
    print(len(res))
    for u,v in res:
        print(u+1,v+1)

import sys

input = lambda :sys.stdin.buffer.readline()
mi = lambda :map(int,input().split())
li = lambda :list(mi())

N,M = mi()
assert  N<= 5000
A = li()
B = li()
E = [tuple(mi()) for i in range(M)]
solve(N,A,B,M,E)
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