【JZOJ5625】Max
Description
Data Constraint
Solution
m
较小,考虑对
再设
预处理
Code
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define fo(i,j,k) for(int i=j;i<=k;++i)
#define fd(i,j,k) for(int i=j;i>=k;--i)
#define ll long long
using namespace std;
const int N=45,M=11,mo=1e9+7;
int p[M][N][4];
int f[N][1<<10][32],g[N][1<<10][32];
int n,m,c;
void add(int &x,int y){
x=x-mo+y;
if(x<0) x+=mo;
}
int calc(int x){
int t=0;
fo(i,1,m) if((1<<i-1)&x) t++;
return t;
}
int max(int x,int y){
return x>y?x:y;
}
int main()
{
freopen("max.in","r",stdin);
freopen("max.out","w",stdout);
scanf("%d %d %d",&n,&m,&c);
fo(i,1,m)
fo(j,1,n)
fo(k,0,c) scanf("%d",&p[i][j][k]);
fo(i,1,n){
g[i][0][0]=1;
fo(s,1,(1<<m)-1){
int x=0,q=0;
fo(j,1,m) if((1<<j-1)&s) {x=s^(1<<j-1),q=j;break;}
fo(k,0,c)
fo(j,0,m*c-k)
add(g[i][s][j+k],(ll)g[i][x][j]*p[q][i][k]%mo);
}
}
f[0][0][0]=1;
fo(i,0,n-1){
int o=(1<<m)-1;
fo(s,0,o){
int S=o^s;
for(int x=S;x>=0;x=(x-1)&S){
int sf=0,sg=0;
fo(j,0,m*c){
add(sg,g[i+1][x][j]);
add(f[i+1][s^x][j],(ll)f[i][s][j]*sg%mo);
add(f[i+1][s^x][j],(ll)g[i+1][x][j]*sf%mo);
add(sf,f[i][s][j]);
}
if(!x) break;
}
}
}
int ans=0;
fo(i,1,m*c) add(ans,(ll)f[n][(1<<m)-1][i]*i%mo);
printf("%d",ans);
}