枚举子集, 把一种颜色的一起加进去dp, 感觉3^n的复杂度不知道为啥跑这么快。
可以用fwt优化到(2 ^ n) * n * n
#pragma GCC optimize(2) #pragma GCC optimize(3) #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0) ; using namespace std; const int N = 3e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = (int)1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;} //mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); int n; bool ok[1 << 18]; char s[18][30]; unsigned int dp[1 << 18]; unsigned int Pow[1 << 18]; int main() { for(int i = Pow[0] = 1; i < (1 << 18); i++) { Pow[i] = Pow[i - 1] * 233; } int T; scanf("%d", &T); while(T--) { scanf("%d", &n); for(int i = 0; i < n; i++) { scanf("%s", s[i]); } for(int i = 0; i < (1 << n); i++) { ok[i] = true; } for(int mask = 0; mask < (1 << n); mask++) { for(int i = 0; i < n; i++) { if(mask >> i & 1) { for(int j = i + 1; j < n; j++) { if(mask >> j & 1) { if(s[i][j] == ‘1‘) { ok[mask] = false; i = n; break; } } } } } } unsigned int ans = 0; dp[0] = 0; for(int i = 1; i < (1 << n); i++) { dp[i] = n; for(int j = i; ; j = (j - 1) & i) { if(ok[i ^ j]) chkmin(dp[i], dp[j] + 1); if(!j) break; } ans += dp[i] * Pow[i]; } printf("%u\n", ans); } return 0; } /* */
原文:https://www.cnblogs.com/CJLHY/p/11438710.html