結果

問題 No.1607 Kth Maximum Card
コンテスト
ユーザー qwewe
提出日時 2025-04-24 12:27:08
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 4,255 bytes
コンパイル時間 245 ms
コンパイル使用メモリ 82,708 KB
実行使用メモリ 225,240 KB
最終ジャッジ日時 2025-04-24 12:28:20
合計ジャッジ時間 12,540 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge1
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 * 7 WA * 26
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()
    idx = 0
    n = int(input[idx]); idx +=1
    m = int(input[idx]); idx +=1
    K = int(input[idx]); idx +=1

    edges = []
    max_c = 0
    for _ in range(m):
        u = int(input[idx]); idx +=1
        v = int(input[idx]); idx +=1
        c = int(input[idx]); idx +=1
        edges.append((u, v, c))
        if c > max_c:
            max_c = c

    # Compute minimal path length using BFS
    adj_shortest = [[] for _ in range(n+1)]
    for u, v, c in edges:
        adj_shortest[u].append(v)
        adj_shortest[v].append(u)
    visited = [False]*(n+1)
    q = deque()
    q.append(1)
    visited[1] = True
    dist = [0]*(n+1)
    while q:
        u = q.popleft()
        if u == n:
            break
        for v in adj_shortest[u]:
            if not visited[v]:
                visited[v] = True
                dist[v] = dist[u] + 1
                q.append(v)
    L = dist[n]
    if L <= K-1:
        print(0)
        return

    # Binary search
    left = 1
    right = max_c
    answer = max_c

    unique_c = sorted(set(c for u, v, c in edges))
    left = 0
    right = len(unique_c) -1

    def is_possible(x_val):
        adj = [[] for _ in range(n+1)]
        for u, v, c in edges:
            if c >= x_val:
                adj[u].append(v)
                adj[v].append(u)
        # Check connectivity
        visited = [False]*(n+1)
        q = deque([1])
        visited[1] = True
        found = False
        while q:
            u = q.popleft()
            if u == n:
                found = True
                break
            for v in adj[u]:
                if not visited[v]:
                    visited[v] = True
                    q.append(v)
        if not found:
            return False
        # Find component containing 1 and N
        component = []
        visited_comp = [False]*(n+1)
        q = deque([1])
        visited_comp[1] = True
        while q:
            u = q.popleft()
            component.append(u)
            for v in adj[u]:
                if not visited_comp[v]:
                    visited_comp[v] = True
                    q.append(v)
        # Check if N is in component
        if not visited_comp[n]:
            return False
        # Check for cycle in the component
        parent = [0]*(n+1)
        has_cycle = False
        visited_cycle = [False]*(n+1)
        for u in component:
            if visited_cycle[u]:
                continue
            q = deque([u])
            parent[u] = -1
            visited_cycle[u] = True
            while q and not has_cycle:
                current = q.popleft()
                for v in adj[current]:
                    if not visited_cycle[v]:
                        visited_cycle[v] = True
                        parent[v] = current
                        q.append(v)
                    else:
                        if parent[current] != v:
                            has_cycle = True
                            break
                if has_cycle:
                    break
        if has_cycle:
            return True
        # Compute longest path (BFS distance in tree)
        distance = [-1]*(n+1)
        q = deque([1])
        distance[1] = 0
        while q:
            u = q.popleft()
            if u == n:
                break
            for v in adj[u]:
                if distance[v] == -1:
                    distance[v] = distance[u] + 1
                    q.append(v)
        return distance[n] >= K

    # Binary search on unique_c
    unique_c.sort()
    left = 0
    right = len(unique_c) -1
    answer = max_c
    while left <= right:
        mid = (left + right) // 2
        x_val = unique_c[mid]
        if is_possible(x_val):
            answer = x_val
            right = mid -1
        else:
            left = mid +1
    # Check if answer is found in unique_c, else check between
    # If not found, there might be a value between unique_c elements
    # So we need to collect all possible c's and binary search accordingly
    # But since the problem requires the K-th largest, which must be one of the c's, we can proceed with unique_c
    print(answer)

if __name__ == '__main__':
    main()
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