#4614. problem B

Title Description

There are two points set $ S, T $, $ \ quad \ forall x \ in S, y \ in T $, $ x, there is $ \ frac {1} {2} $ probability that an edge exists between y $.

Now from any $ S, T $ each randomly pick a point, ask a desired distance between the two points.

If none, then the distance is $ 0 $

Output at a desired distance value $ P $ domain mold, to ensure that $ P $ is a prime number.

data range

For $ 100 \ $% of the data, to ensure that $ 1 \ le n, m \ le 100, 772001 \ le p \ le 1000000007 $, to ensure that p is a prime number.

answer

Autistic exam questions.

Violence may be seen as if the sum of the shortest went 1 $ $ $ T $ point number in / from program number 1 $ S $ $ $ point number, i.e. for each program to $ $ BFS again.

$ $ BFS process consideration, because it is a bipartite graph, so the first layer $ i $ only point and the point of $ i-1 $ a linked layer side.

Therefore, with the process of $ DP $ analog $ BFS $, the set $ f_ {i, j, k, 0/1} $ represents $ S $ selected a $ I $ points, $ T $ selected a $ J $ points, the last layer has $ k $ points, which points are $ k $ $ S / T $ program number, set $ G $ is the sum of the depth of the deepest of all programs.

When the transfer of the election does not determine what option $ 1 $ dot $ T $ in the way statistics can answer.

Efficiency: $ O (n ^ 4) $, note cards often.

Code

#include <bits/stdc++.h>
using namespace std; const int N=105;
int n,m,P,f[N][N][N][2],g[N][N][N][2],p[N][N],c[N][N],s,h[N*N],d[N][N][N][2];
inline int X(int x){return x>=P?x-P:x;}
int K(int x,int y){
    int z=1;
    for (;y;y>>=1,x=1ll*x*x%P)
        if (y&1) z=1ll*z*x%P;
    return z;
}
int main(){
    scanf("%d%d%d",&n,&m,&P);if (n<m) swap(n,m);
    c[0][0]=f[1][0][1][0]=h[0]=1;
    for (int i=1;i<=n;i++){
        c[i][0]=1;
        for (int j=1;j<=i;j++)
            c[i][j]=X(c[i-1][j-1]+c[i-1][j]);
    }
    for (int i=1;i<=n;i++){
        p[i][1]=X(X(p[i-1][1]<<1)+1);
        for (int j=2;j<=n;j++)
            p[i][j]=1ll*p[i][j-1]*p[i][1]%P;
    }
    for (int i=0;i<=n;i++)
        for (int j=0;j<=n;j++)
            for (int k=0;k<=n;k++)
                d[i][j][k][0]=1ll*c[i][k]*p[j][k]%P,
                d[i][j][k][1]=1ll*c[i][k]*p[j][k+1]%P;
    for (int i=1;i<=n*m;i++) h[i]=X(h[i-1]<<1);m--;
    for (int i=1;i<=n;i++)
        for (int j=0;j<=m;j++){
            for (int w,k=1;k<=i;++k){
                w=p[k][1];
                for (int x=1;x+j<=m;++x)
                    w=X(w+d[m-j][k][x][1]),
                    f[i][j+x][x][1]=X(f[i][j+x][x][1]+1ll*d[m-j][k][x][0]*f[i][j][k][0]%P),
                    g[i][j+x][x][1]=X(g[i][j+x][x][1]+1ll*d[m-j][k][x][0]*(g[i][j][k][0]+f[i][j][k][0])%P);
                s=X(s+1ll*w*h[(n-i)*(m-j+1)]%P*(g[i][j][k][0]+f[i][j][k][0])%P);
            }
            for (int k=1;k<=j;++k)
                for (int x=1;x+i<=n;++x)
                    f[i+x][j][x][0]=X(f[i+x][j][x][0]+1ll*d[n-i][k][x][0]*f[i][j][k][1]%P),
                    g[i+x][j][x][0]=X(g[i+x][j][x][0]+1ll*d[n-i][k][x][0]*(g[i][j][k][1]+f[i][j][k][1])%P);
        }
    printf("%lld\n",1ll*s*K(h[n*(m+1)],P-2)%P);return 0;
}