解题思路:
ovo
参考代码:
#include <bits/stdc++.h> constexpr auto Inf = 0x3F3F3F3F; typedef long long LL; using namespace std; namespace IO { inline LL read() { LL o = 0, f = 1; char c = getchar(); while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); } while (c > '/' && c < ':') { o = o * 10 + c - '0'; c = getchar(); } return o * f; } inline char recd() { char o; while ((o = getchar()) != 'C' && o != 'P'); return o; } } using namespace IO; const int SIZE = 17; int Arr[SIZE][SIZE], F[SIZE], Dp[SIZE][1 << SIZE], Ind[SIZE], M; int BIT(int pos) { int ans = 0; while (pos) pos >>= 1, ans++; return ans; } int main() { scanf("%d", &M); for (int pos = 0; pos < M; pos++) F[pos] = 1 << pos, Ind[pos] = BIT(F[pos]); for (int posx = 1; posx <= M; posx++) for (int posy = 1; posy <= M; posy++) scanf("%d", &Arr[posx][posy]); for (int pos = 0; pos < M; pos++) Dp[1][F[pos]] = Arr[1][Ind[pos]]; for (int pos = 2; pos <= M; pos++) for (int Status = 0; Status < M; Status++) for (int now = 1; now < 1 << M; now++) if (!(now & F[Status])) Dp[pos][F[Status] ^ now] = max(Dp[pos][F[Status] ^ now], Dp[pos - 1][now] + Arr[pos][Ind[Status]]); printf("%d\n", Dp[M][(1 << M) - 1]); }
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