結果

問題 No.1341 真ん中を入れ替えて門松列
コンテスト
ユーザー chineristAC
提出日時 2021-01-15 22:16:30
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 2,789 bytes
コンパイル時間 148 ms
コンパイル使用メモリ 82,564 KB
実行使用メモリ 247,908 KB
最終ジャッジ日時 2024-11-26 15:50:40
合計ジャッジ時間 35,844 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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 * 5
other AC * 3 TLE * 11
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 # 

from heapq import heappush, heappop
class MinCostFlow:
    INF = 10**18

    def __init__(self, N):
        self.N = N
        self.G = [[] for i in range(N)]

    def add_edge(self, fr, to, cap, cost):
        forward = [to, cap, cost, None]
        backward = forward[3] = [fr, 0, -cost, forward]
        self.G[fr].append(forward)
        self.G[to].append(backward)

    def flow(self, s, t, f):
        N = self.N; G = self.G
        INF = MinCostFlow.INF

        res = 0
        H = [0]*N
        prv_v = [0]*N
        prv_e = [None]*N

        d0 = [INF]*N
        dist = [INF]*N

        while f:
            dist[:] = d0
            dist[s] = 0
            que = [(0, s)]

            while que:
                c, v = heappop(que)
                if dist[v] < c:
                    continue
                r0 = dist[v] + H[v]
                for e in G[v]:
                    w, cap, cost, _ = e
                    if cap > 0 and r0 + cost - H[w] < dist[w]:
                        dist[w] = r = r0 + cost - H[w]
                        prv_v[w] = v; prv_e[w] = e
                        heappush(que, (r, w))
            if dist[t] == INF:
                return None

            for i in range(N):
                H[i] += dist[i]

            d = f; v = t
            while v != s:
                d = min(d, prv_e[v][1])
                v = prv_v[v]
            f -= d
            res += d * H[t]
            v = t
            while v != s:
                e = prv_e[v]
                e[1] -= d
                e[3][1] += d
                v = prv_v[v]
        return res

N,M = map(int,input().split())

AC = []
B = []
for i in range(N):
    a,b,c = map(int,input().split())
    AC.append((a,c))
    B.append(b)

G = MinCostFlow(4*N+2)

for i in range(N):
    G.add_edge(0,i+1,1,0)

idx = [i for i in range(N)]
idx.sort(key=lambda i:min(AC[i]))
for _ in range(N-1):
    pos = idx[_]
    next = idx[_+1]
    G.add_edge(N+1+pos,N+1+next,10**15,0)
for _ in range(N):
    pos = idx[_]
    G.add_edge(N+1+pos,3*N+1+pos,1,-max(AC[pos]))
for i in range(N):
    b = B[i]
    for j in range(N):
        if min(AC[idx[j]]) > b:
            G.add_edge(i+1,N+1+idx[j],1,0)
            break

idx.sort(key=lambda i:max(AC[i]))
for _ in range(N-1,0,-1):
    pos = idx[_]
    next = idx[_-1]
    G.add_edge(2*N+1+pos,2*N+1+next,10**15,0)
for i in range(N):
    b = B[i]
    for j in range(N-1,-1,-1):
        if max(AC[idx[j]]) < b:
            G.add_edge(i+1,2*N+1+idx[j],1,-b)
            break

for i in range(N):
    G.add_edge(N+1+i,3*N+1+i,1,0)
    G.add_edge(2*N+1+i,3*N+1+i,1,0)
    G.add_edge(3*N+1+i,4*N+1,1,0)

res = G.flow(0,4*N+1,N)
if res:
    print("YES")
    res = - res
    if res >= M:
        print("KADOMATSU!")
    else:
        print("NO")
else:
    print("NO")
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