結果

問題 No.1265 Balloon Survival
ユーザー lam6er
提出日時 2025-04-15 23:22:34
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 3,434 bytes
コンパイル時間 364 ms
コンパイル使用メモリ 82,156 KB
実行使用メモリ 97,276 KB
最終ジャッジ日時 2025-04-15 23:24:08
合計ジャッジ時間 6,875 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
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ファイルパターン 結果
sample AC * 4
other AC * 9 WA * 23
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ソースコード

diff # 

import math
from collections import deque

def main():
    import sys
    input = sys.stdin.read().split()
    idx = 0
    N = int(input[idx])
    idx +=1
    x0 = int(input[idx])
    y0 = int(input[idx+1])
    idx +=2
    balloons = []
    for _ in range(N-1):
        x = int(input[idx])
        y = int(input[idx+1])
        idx +=2
        balloons.append((x, y))
    
    if N == 1:
        print(0)
        return
    
    # Compute t_i for each balloon
    t = []
    for (x, y) in balloons:
        dx = x - x0
        dy = y - y0
        dist = math.hypot(dx, dy)
        t_i = dist / 2.0
        t.append(t_i)
    
    # Build adjacency list for bipartite graph
    adj = [[] for _ in range(len(balloons))]
    for i in range(len(balloons)):
        xi, yi = balloons[i]
        for j in range(i+1, len(balloons)):
            xj, yj = balloons[j]
            dx = xi - xj
            dy = yi - yj
            dist_ij = math.hypot(dx, dy)
            t_ij = dist_ij / 2.0
            if t_ij < min(t[i], t[j]):
                adj[i].append(j)
                adj[j].append(i)
    
    # Bipartite graph: split each node into left and right
    # Use Hopcroft-Karp algorithm
    # Each node is represented in left and right partitions
    # So node i in left is i, in right is len(balloons) + i
    size_left = len(balloons)
    size_right = len(balloons)
    graph = [[] for _ in range(size_left + size_right)]
    for i in range(len(balloons)):
        for j in adj[i]:
            graph[i].append(len(balloons) + j)
            graph[len(balloons) + j].append(i)
    
    # Hopcroft-Karp algorithm
    def hopcroft_karp():
        pair_U = [-1] * (size_left + size_right)
        pair_V = [-1] * (size_left + size_right)
        dist = [0] * (size_left + size_right)
        
        def bfs():
            queue = deque()
            for u in range(size_left):
                if pair_U[u] == -1:
                    dist[u] = 0
                    queue.append(u)
                else:
                    dist[u] = float('inf')
            dist_null = float('inf')
            while queue:
                u = queue.popleft()
                if dist[u] < dist_null:
                    for v in graph[u]:
                        if pair_V[v] == -1:
                            dist_null = dist[u] + 1
                        elif dist[pair_V[v]] == float('inf'):
                            dist[pair_V[v]] = dist[u] + 1
                            queue.append(pair_V[v])
            return dist_null != float('inf')
        
        def dfs(u):
            for v in graph[u]:
                if pair_V[v] == -1 or (dist[pair_V[v]] == dist[u] + 1 and dfs(pair_V[v])):
                    pair_U[u] = v
                    pair_V[v] = u
                    return True
            dist[u] = float('inf')
            return False
        
        result = 0
        while bfs():
            for u in range(size_left):
                if pair_U[u] == -1:
                    if dfs(u):
                        result +=1
        return result
    
    max_matching = hopcroft_karp()
    # Each pair in bipartite graph corresponds to a pair in original graph, but each original edge is represented twice
    # So the actual maximum matching in the original graph is max_matching // 2
    actual_matching = max_matching // 2
    answer = (N-1) - 2 * actual_matching
    print(answer)

if __name__ == "__main__":
    main()
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