26. Lost in WHU--武汉大学网络赛

标题 26. Lost in WHU

Input file: standard input
Output file: standard output
Time limit: 1 second
Memory limit: 512 mebibytes

As one of the most beautiful campus in China, Wuhan University is around several hills, so the road is complex and visitors always lose themselves. Give a undirected graph of WHU of NN points and a deadline, then you should count the number of plan to reach the destination by deadline (the starting point is 1 and the ending point is NN).

Input

First line contains the number of points NN (N\le 100N≤100) and the number of edges MM (M\le N(N-1)/2M≤N(N−1)/2). The ii-th line of next MM lines contains two numbers u_iu ​i ​​ and v_iv ​i ​​ , which respents an edge (u_i, v_i)(u ​i ​​ ,v ​i ​​ ). The last line contains the deadline TT(T\le 10^9T≤10 ​9 ​​ ).

Output
The number of plan to reach the destination by deadline module 10^9+710 ​9 ​​ +7.

Input 1
4 5
1 3
2 3
3 4
1 2
1 4
8
Output 1
170

题目大意：有n个点，m条边，每条边代表这条路能走，有k时间。
问：从1到n，有多少种走法；
思路 1：可以往回走，只要没有到达n和时间没有结束；
         2：到达n后不能再走了，即使时间未到；
         3：时间未到时，到达n，也算一次；

#include
#define mod 1000000007
struct mat
{
    long long a[110][110];
};
mat I;
int n;
mat mat_mul(mat q,mat w)
{
    int i,j;
    mat A;
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=n; j++)
        {
            A.a[i][j]=0;
            for(int k=1; k<=n; k++)
            {
                A.a[i][j]+=(q.a[i][k]*w.a[k][j])%mod;//结构体，要开long long,乘的时候会爆表；
                A.a[i][j]%=mod;
            }
        }
    }
    return A;
}
mat mul(mat q,long long k)
{
    mat sum;
    for(int i=1; i//初始化，单位矩阵；
        sum.a[i][i]=1;
    mat c=q;
    while(k)
    {
        if(k&1)
            sum=mat_mul(sum,c);
        c=mat_mul(c,c);
        k>>=1;
    }
    return sum;
}
int main()
{
    int m;
    long long k;
    while(~scanf("%d%d",&n,&m))
    {
        mat map,ans;
        int i,j;
        for(i=0; iint x,y;
            scanf("%d%d",&x,&y);
            if(x!=n) map.a[x][y]=1;//保证到达n时，不再走了；
            if(y!=n) map.a[y][x]=1;
        }
        map.a[n][n]=1;//重点；
        //关键：与一般矩阵快速幂不同，这一步可以加上时间未到，但到达n              的也可以加起来；
        //一般的矩阵快速幂是到n时恰好是k步的有多少步；
        scanf("%I64d",&k);
        ans=mul(map,k);
        printf("%I64d\n",ans.a[1][n]);
    }
    return 0;
}