HDU 5117 Fluorescent (数学+状压DP)

题意：有 n 个灯,初始状态都是关闭，有m个开关，每个开关都控制若干个。问在m个开关按下与否的2^m的情况中，求每种情况下亮灯数量的立方和。

析：首先，如果直接做的话，时间复杂度无法接受，所以要对其进行小小的变形，设开灯数X，和每个开关的状态的对应关系是X = x1+x2+...+xn，其中 xi 可能为0，可能为1，那么要求的数X^3 = (x1+x2+...+xn) * (x1+x2+...+xn) * (x1+x2+...+xn) = ∑(xi*xj*xk)，只有xi = xj = xk时，那么这种情况才会成立，所以对xi,xj,xk 进行枚举，dp[t][s] 表示前 t 个开关时，状态为s的可能数，这个状态 s 表示的是 xi,xj,xk的状态，只有s == 7 时，才满足条件。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50 + 50;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

LL dp[maxn][8];
LL st[maxn];

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= m; ++i){
      st[i] = 0;
      int x, y;
      scanf("%d", &y);
      for(int j = 0; j < y; ++j){
        scanf("%d", &x);
        st[i] |= 1LL<<x-1;
      }
    }
    LL ans = 0;
    FOR(i, 0, n)  FOR(j, 0, n)  FOR(k, 0, n){
      ms(dp, 0);  dp[0][0] = 1;
      for(int t = 1; t <= m; ++t){
        for(int s = 0; s < 8; ++s){
          int newst = 0;
          if(st[t]&1LL<<i)  newst |= 1;
          if(st[t]&1LL<<j)  newst |= 2;
          if(st[t]&1LL<<k)  newst |= 4;
          dp[t][s] = dp[t-1][s] + dp[t-1][s^newst];
        }
      }
      ans = (ans + dp[m][7]) % mod;
    }
    printf("Case #%d: %I64d\n", kase, ans);
  }
  return 0;
}