結果

問題 No.2915 辺更新価値最大化
コンテスト
ユーザー Shirotsume
提出日時 2024-07-29 19:41:38
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 1,996 bytes
コンパイル時間 304 ms
コンパイル使用メモリ 81,988 KB
実行使用メモリ 87,472 KB
最終ジャッジ日時 2024-07-29 20:31:12
合計ジャッジ時間 10,017 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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ファイルパターン 結果
other AC * 25 TLE * 1 -- * 2
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, time, random
from collections import deque, Counter, defaultdict
input = lambda: sys.stdin.readline().rstrip()
ii = lambda: int(input())
mi = lambda: map(int, input().split())
li = lambda: list(mi())
inf = 2 ** 61 - 1
mod = 998244353
def Dijkstra(s, graph):
    INF = 2 ** 61 - 1
    import heapq
    n = len(graph)
    dist = [INF] * n
    dist[s] = 0
    bef = [0] * n
    bef[s] = s
    hq = [(0, s)]
    heapq.heapify(hq)
    while hq:
        c, now = heapq.heappop(hq)
        
        if c > dist[now]:
            continue
        for to, cost in graph[now]:
            if dist[now] + cost < dist[to]:
                dist[to] = cost + dist[now]
                bef[to] = now
                heapq.heappush(hq, (dist[to], + to))
    return dist, bef
n, m, q = mi()
assert 1 <= n <= 10 ** 3 and 1 <= m <= 10 ** 3 and 1 <= q <= 10 ** 3
graph = [[] for _ in range(n)]
EDGE = []
for _ in range(m):
    u, v, c = mi()
    assert 1 <= u <= n and 1 <= v <= n and -10 ** 3 <= c <= 10 ** 3
    assert u != v
    u -= 1
    v -= 1
    c *= -1
    EDGE.append((u, v, c))
    graph[u].append((v, c))
    
#ベルマンフォード法
dist = [inf] * n
dist[0] = 0
for _ in range(n):
    for j in range(n):
        for to, c in graph[j]:
            if dist[to] > dist[j] + c:
                dist[to] = dist[j] + c

for i in range(n):
    for to, c in graph[j]:
        assert dist[to] <= dist[j] + c
graph2 = [[] for _ in range(n)]
EDGE2 = []
d1 = dist[n - 1]
for u, v, c in EDGE:
    EDGE2.append((u, v, c + dist[v] - dist[u]))
    graph2[u].append((v, c + dist[v] - dist[u]))
onoff = [1] * m
E = li()
for e in E:
    assert 1 <= e <= m
    e -= 1
    if onoff[e]:
        onoff[e] = 0
        graph2[EDGE2[e][0]].remove((EDGE2[e][1], EDGE2[e][2]))
    else:
        onoff[e] = 1
        graph2[EDGE2[e][0]].append((EDGE2[e][1], EDGE2[e][2]))
    dist, bef = Dijkstra(0, graph2)
    if dist[n - 1] == inf:
        print('NaN')
    else:
        print(-(dist[n - 1] - d1))
0