BZOJ4887 Tjoi2017可乐（动态规划+矩阵快速幂）

　　设f[i][j]为第i天到达j号城市的方案数，转移显然，答案即为每天在每个点的方案数之和。矩乘一发即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 33
#define M 110
#define P 2017
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,m,k;
struct matrix
{
    int n,a[N][N];
    matrix operator *(const matrix&b) const
    {
        matrix c;c.n=n;memset(c.a,,sizeof(c.a));
        for (int i=;i<n;i++)
            for (int j=;j<b.n;j++)
                for (int k=;k<b.n;k++)
                c.a[i][j]=(c.a[i][j]+a[i][k]*b.a[k][j])%P;
        return c;
    }
}f,v;
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4887.in","r",stdin);
    freopen("bzoj4887.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    v.n=n+;
    for (int i=;i<=m;i++)
    {
        int x=read(),y=read();
        v.a[x][y]=v.a[y][x]=;
    }
    for (int i=;i<=n;i++) v.a[i][i]=v.a[i][]=;
    k=read()+;
    f.n=;f.a[][]=;
    for (;k;k>>=,v=v*v) if (k&) f=f*v;
    cout<<f.a[][];
    return ;
}