题意：给定一张n点m边的图，点带点权，定义点覆盖的权值为点权之积，问所有点覆盖的权值之和膜q
n<=36， 1<=a[i]<=1e9，1e8<=q<=1e9
思路：n<=36，考虑middle in the middle分成两个点数接近的点集L和R
对于L，枚举其子集S，判断S能否覆盖所有L内部的边，预处理出所有合法的S的超集的贡献
对于R，枚举其子集T，判断T能否覆盖所有R内部的边，如果可以则可以推出L，R之间在确定R中选T的前提下左边至少需要选点集T’，答案即为T的点权之积*T’的超集的点权积之和
对于判断覆盖和根据T推T'使用了大量位运算加速
需要注意的是如果二进制左右移位可能超边界则要使用ull
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef pair<int,int> PII; 7 typedef pair<ll,ll> Pll; 8 typedef vector<int> VI; 9 typedef vector<PII> VII; 10 //typedef pair<ll,ll>P; 11 #define N 300010 12 #define M 2000010 13 #define fi first 14 #define se second 15 #define MP make_pair 16 #define pb push_back 17 #define pi acos(-1) 18 #define mem(a,b) memset(a,b,sizeof(a)) 19 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 20 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 21 #define lowbit(x) x&(-x) 22 #define Rand (rand()*(1<<16)+rand()) 23 #define id(x) ((x)<=B?(x):m-n/(x)+1) 24 #define ls p<<1 25 #define rs p<<1|1 26 27 const //ll MOD=1e9+7,inv2=(MOD+1)/2; 28 double eps=1e-6; 29 int INF=1e9; 30 int dx[4]={-1,1,0,0}; 31 int dy[4]={0,0,-1,1}; 32 33 ull s[M]; 34 ll a[N],f[N]; 35 36 int read() 37 { 38 int v=0,f=1; 39 char c=getchar(); 40 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 41 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 42 return v*f; 43 } 44 45 int isok1(int S,int l,int r) 46 { 47 rep(i,l,r) 48 if(!(S>>i&1)) 49 { 50 ull now=((s[i]<<(63-r))>>(63-r)); 51 if((now&S)!=now) return 0; 52 } 53 return 1; 54 } 55 56 int isok2(int S,int l,int r,int mid) 57 { 58 rep(i,l,r) 59 if(!(S>>i&1)) 60 { 61 ll now=(s[i+mid]>>mid); 62 if((now&S)!=now) return 0; 63 } 64 return 1; 65 } 66 67 int main() 68 { 69 int cas=read(); 70 rep(v,1,cas) 71 { 72 int n=read(),m=read(); 73 ll MOD; 74 scanf("%I64d",&MOD); 75 int mid=n/2; 76 rep(i,0,n-1) scanf("%I64d",&a[i]); 77 mem(s,0); 78 rep(i,1,m) 79 { 80 int x=read(),y=read(); 81 x--; y--; 82 s[x]|=1ll<<y; 83 s[y]|=1ll<<x; 84 } 85 int S1=(1<<mid)-1; 86 rep(i,0,S1) f[i]=0; 87 rep(i,0,S1) 88 { 89 ll t=1; 90 rep(j,0,mid-1) 91 if(i>>j&1) t=t*a[j]%MOD; 92 if(isok1(i,0,mid-1)) f[i]=t; 93 } 94 95 rep(i,0,mid-1) 96 rep(j,0,S1) 97 if(!(j>>i&1)) f[j]=(f[j]+f[j^(1<<i)])%MOD; 98 99 int S2=(1<<(n-mid))-1; 100 ll ans=0; 101 rep(i,0,S2) 102 { 103 ll t=1; 104 rep(j,0,n-mid-1) 105 if(i>>j&1) t=t*a[j+mid]%MOD; 106 if(isok2(i,0,n-mid-1,mid)) 107 { 108 ll base=0; 109 rep(j,0,mid-1) 110 { 111 ull now=(s[j]>>mid); 112 if((now&i)!=now) base|=1<<j; 113 } 114 ans=(ans+t*f[base]%MOD)%MOD; 115 } 116 } 117 printf("Case #%d: ",v); 118 printf("%I64d ",ans); 119 } 120 return 0; 121 }