結果

問題 No.1612 I hate Construct a Palindrome
ユーザー gew1fw
提出日時 2025-06-12 16:43:54
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 4,324 bytes
コンパイル時間 704 ms
コンパイル使用メモリ 82,360 KB
実行使用メモリ 171,892 KB
最終ジャッジ日時 2025-06-12 16:44:14
合計ジャッジ時間 9,474 ms
ジャッジサーバーID
（参考情報） 		judge3 / 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 * 3
other AC * 26 WA * 10
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 # 

import sys
from collections import deque

def main():
    input = sys.stdin.read().split()
    ptr = 0
    N = int(input[ptr])
    ptr += 1
    M = int(input[ptr])
    ptr += 1

    edges = []
    adj = [[] for _ in range(N+1)]  # 1-based
    for i in range(M):
        A = int(input[ptr])
        ptr += 1
        B = int(input[ptr])
        ptr += 1
        C = input[ptr]
        ptr += 1
        edges.append((A, B, C))
        adj[A].append((B, C, i))
        adj[B].append((A, C, i))

    # BFS from 1 to compute parent_1
    parent_1 = [-1] * (N + 1)
    visited = [False] * (N + 1)
    q = deque([1])
    visited[1] = True

    while q:
        u = q.popleft()
        for v, c, idx in adj[u]:
            if not visited[v]:
                visited[v] = True
                parent_1[v] = idx
                q.append(v)

    # Reconstruct the shortest path from 1 to N
    path_1_to_N = []
    current = N
    if not visited[N]:
        print(-1)
        return

    while current != 1:
        edge_idx = parent_1[current]
        if edge_idx == -1:
            print(-1)
            return
        path_1_to_N.append(edge_idx)
        a, b, c = edges[edge_idx]
        if a == current:
            current = b
        else:
            current = a
    path_1_to_N.reverse()

    if not path_1_to_N:
        print(-1)
        return

    first_edge_idx = path_1_to_N[0]
    last_edge_idx = path_1_to_N[-1]
    first_char = edges[first_edge_idx][2]
    last_char = edges[last_edge_idx][2]

    if first_char != last_char:
        print(len(path_1_to_N))
        for idx in path_1_to_N:
            print(idx + 1)
        return

    # Need to find another path
    c = first_char

    # Check edges from 1 with label != c
    found = False
    for v, c_label, idx in adj[1]:
        if c_label != c:
            # Reconstruct path from v to 1
            path_v_to_1 = []
            current = v
            valid = True
            while current != 1:
                edge_idx = parent_1[current]
                if edge_idx == -1:
                    valid = False
                    break
                path_v_to_1.append(edge_idx)
                a, b, c_e = edges[edge_idx]
                if a == current:
                    current = b
                else:
                    current = a
            if not valid:
                continue

            # Reconstruct path from 1 to N again
            path_1_to_N_current = []
            current = N
            while current != 1:
                edge_idx = parent_1[current]
                path_1_to_N_current.append(edge_idx)
                a, b, c_e = edges[edge_idx]
                if a == current:
                    current = b
                else:
                    current = a
            path_1_to_N_current.reverse()

            full_path = [idx] + path_v_to_1 + path_1_to_N_current
            if len(full_path) > 2 * N:
                continue

            print(len(full_path))
            for p in full_path:
                print(p + 1)
            return

    # Check edges to N with label != c
    for v, c_label, idx in adj[N]:
        if c_label != c:
            # Reconstruct path from 1 to v
            path_1_to_v = []
            current = v
            valid = True
            while current != 1:
                edge_idx = parent_1[current]
                if edge_idx == -1:
                    valid = False
                    break
                path_1_to_v.append(edge_idx)
                a, b, c_e = edges[edge_idx]
                if a == current:
                    current = b
                else:
                    current = a
            if not valid:
                continue
            path_1_to_v.reverse()

            # Check first edge of path_1_to_v
            if not path_1_to_v:
                first_char_v = c_label  # path is 1 to v directly via edge from 1 to v?
            else:
                first_char_v = edges[path_1_to_v[0]][2]

            if first_char_v != c_label:
                full_path = path_1_to_v + [idx]
                if len(full_path) > 2 * N:
                    continue
                print(len(full_path))
                for p in full_path:
                    print(p + 1)
                return

    print(-1)

if __name__ == "__main__":
    main()
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