[HDU - 5117 ]-Fluorescent （状态压缩DP，期望）
题目链接： HDU - 5117
题面：
题意：
给定(mathit n)个灯泡，初始全是熄灭状态。
(mathit m)个开关，每一个开关可以控制一些灯。当该开关被按下时，它控制的灯泡状态将取反。
每一个开关将被等概率的按下/不按下。设(mathit X) 为最终被点亮的灯泡个数，求(E[X^3] × 2^M mod (10^9 + 7))
思路：
[E[X^3] imes2^M=sum (X^{3} * (frac{1}{2})^{M}) * 2^{M} = sum X^{3}
=sum_{i,j,m=1}^{n}[l_i*l_j*l_k]
]
于是问题转化为，枚举所有的(i,j,k)，求有多少种方案数使其让这三个灯都亮着，所有情况的总和就是答案。
我们对每一个((i,j,k))三元组进行状压DP求解方案数，
设(dp[s][w]) 代表对于前(1dots m) 个开关，使其（i,j,k）三个灯是否点亮的状态为s的方案数。
初始状态：(dp[0][0]=1)。
转移方程：对于第(mathit w)个开关，（i,j,k）三个灯是否被其控制的状态为(mathit s)。
那么(dp[q][w]=dp[q][w-1]+dp[qoplus s][w-1],qin[1,7])。
该三元组对答案的贡献为(dp[(111)_2][m])。
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '
' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '
' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
const ll mod = 1e9 + 7;
int dp[8][55];
int n, m;
ll info[55];
int main()
{
#if DEBUG_Switch
freopen("D:\code\input.txt", "r", stdin);
#endif
//freopen("D:\code\output.txt","w",stdout);
int t;
t = readint();
for (int icase = 1; icase <= t; ++icase) {
n = readint(), m = readint();
repd(i, 1, m) {
info[i] = 0ll;
int num = readint();
while (num--) {
int x = readint();
info[i] |= (1ll << x);
}
}
ll ans = 0ll;
repd(i, 1, n) {
repd(j, 1, n) {
repd(k, 1, n) {
MS0(dp);
dp[0][0] = 1ll;
repd(w, 1, m) {
int s = 0;
if (info[w] & (1ll << i)) { s |= 1; }
if (info[w] & (1ll << j)) { s |= 2; }
if (info[w] & (1ll << k)) { s |= 4; }
repd(q, 0, 7) {
dp[q][w] = (dp[s ^ q][w - 1] + dp[q][w - 1]) % mod;
}
}
ans = (ans + dp[7][m]) % mod;
}
}
}
printf("Case #%d: %lld
", icase, ans );
}
return 0;
}