HDU5117 - Fluorescent 【状压DP】
化简 Ex3 。X = x1+x2+x3……xn
X3 = ∑xixjxk
即求i,j,k全亮的个数
三层循环I,j,k 对每一个i,j,k 进行DP
t代表第几个开关,x代表状态,x从0到7枚举。
s是状态改变
按:DP[t+1][x^s]+=DP[t][x]
不按:DP[t+1][x]+=DP[t][x]
ans+=DP[m][7]
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 #define prt(k) cout<<#k" = "<<k<<endl; 6 typedef long long ll; 7 #include <algorithm> 8 9 const ll mod = 1e9 + 7; 10 const int N = 52; 11 ll S[N]; 12 int DP[N][8]; 13 int n,m ; 14 int i, j, k; 15 int g( ll s) 16 { 17 int ret = 0; 18 if (s >> i & 1) ret ^= 1; 19 if (s >> j & 1) ret ^= 2; 20 if (s >> k & 1) ret ^= 4; 21 return ret; 22 } 23 void add(int &a, int b) { a=(a+b)%mod; } 24 int main() 25 { 26 int re; int ca=1; 27 cin>>re; 28 while (re--) 29 { 30 cin>>n>>m; 31 for (int i=0;i<m;i++) 32 { 33 ll t = 0; 34 int k; cin>>k; 35 while (k--) 36 { 37 int x ;scanf("%d", &x); 38 x--; 39 t |= (1ll<<x); 40 } 41 S[i] = t; 42 } 43 int ans = 0; 44 for (i=0;i<n;i++) 45 { 46 for (j=0;j<n;j++) 47 { 48 for (k=0;k<n;k++) 49 { 50 memset(DP, 0, sizeof DP); 51 DP[0][0] = 1; 52 for (int t=0;t<m;t++) 53 { 54 int s = g(S[t]); 55 for (int x=0;x<8;x++) 56 { 57 add(DP[t+1][x^s], DP[t][x]); 58 add(DP[t+1][x], DP[t][x]); 59 } 60 } 61 add(ans, DP[m][7]); 62 } 63 } 64 } 65 printf("Case #%d: %d\n", ca++, ans); 66 } 67 return 0; 68 }