結果

問題 No.1078 I love Matrix Construction
ユーザー gr1msl3y
提出日時 2022-01-28 00:10:59
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,729 ms / 2,000 ms
コード長 2,025 bytes
コンパイル時間 399 ms
コンパイル使用メモリ 82,176 KB
実行使用メモリ 268,308 KB
最終ジャッジ日時 2024-12-26 09:10:08
合計ジャッジ時間 17,821 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
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ファイルパターン 結果
other AC * 22
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ソースコード

diff #
プレゼンテーションモードにする

from collections import deque, defaultdict
def scc(N, G, RG):
    order = []
    seen = [0]*N
    group = [None]*N
    def dfs(s):
        task = deque([(s, 1), (s, 0)])
        while task:
            v, t = task.pop()
            if t:
                if seen[v] == 2:
                    continue
                seen[v] = 2
                order.append(v)
            else:
                if seen[v]:
                    continue
                seen[v] = 1
                for e in G[v]:
                    if not seen[e]:
                        task.append((e, 1))
                        task.append((e, 0))
    def rdfs(s, col):
        task = deque([s])
        while task:
            v = task.pop()
            if seen[v]:
                continue
            seen[v] = 1
            group[v] = col
            for e in RG[v]:
                if not seen[e]:
                    task.append(e)
    for i in range(N):
        if not seen[i]:
            dfs(i)
    seen = [0]*N
    label = 0
    for s in reversed(order):
        if not seen[s]:
            rdfs(s, label)
            label += 1
    return label, group
N = int(input())
S = list(map(lambda x: int(x)-1, input().split()))
T = list(map(lambda x: int(x)-1, input().split()))
U = list(map(int, input().split()))
graph = defaultdict(list)
rgraph = defaultdict(list)
for i in range(N):
    for j in range(N):
        s, t, u = S[i], T[i], U[i]
        x = s*N+j
        y = j*N+t
        nx = x+N**2*(u % 2)
        ny = y+N**2*(u//2)
        graph[nx].append((ny+N**2) % (2*N**2))
        rgraph[(ny+N**2) % (2*N**2)].append(nx)
        graph[ny].append((nx+N**2) % (2*N**2))
        rgraph[(nx+N**2) % (2*N**2)].append(ny)
M, sccgraph = scc(2*N**2, graph, rgraph)
ans = [[0]*N for _ in range(N)]
for i in range(N):
    for j in range(N):
        n = i*N+j
        if sccgraph[n] == sccgraph[n+N**2]:
            print(-1)
            exit()
        elif sccgraph[n] < sccgraph[n+N**2]:
            ans[i][j] = 1
for a in ans:
    print(*a)
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