Found

Found

Time Limit: 10000ms, Special Time Limit:25000ms, Memory Limit:260144KB

Total submit users: 2, Accepted users: 2

Problem 13881 : No special judgement

Problem description

Alice and Bob live in a city with N intersections (numbered as intersection 1, intersection 2,…, intersection N). There are M roads. Each road is bidirectional and directly joins two intersections. Each pair of intersections is joined by at most one road. Alice lives in a house at intersection 1. Bob lives in a house at intersection N. They start wandering in the city at 12pm. Each minute each one move from one intersection to another which is connected directly by a road. (Note that since they have to move, they cannot stay at the same intersection they were at in the previous minute). They agree to meet exactly T minutes after 12pm at some intersection. Also they do not want to meet each other before T minutes. In how many ways can they move inside the city? Note that because they travel fairly fast, they do not meet if they travel on the same road on different directions.

Input

The first line of the input contains an integer K (1<=K<=5), the number of test cases. Then K test cases follow in the format specified below. Each test case starts with 3 integers N, M, and T (2<=N<=10; 1<=M<=50; 1<=T<=1,000,000,000,000).  The next M lines specify road connections. That is, line 1+i, for 1<=i<=M, contain two integers A and B specifying that there is a bidirectional road connecting intersection A and B (1<=A<=N; 1<=B<=N; A is not equal to B).

Output

For each test case, your program should output the number of possible ways (modulo 9973) can Alice and Bob move in the city so that at time T they meet at some intersection but they do not meet at any intersection before that. Since the number can be very large, you should output the numbers modulo 9973.

Sample Input

4 3 3 2 1 2 2 3 1 3 3 2 2 1 2 2 3 4 4 10 1 2 2 4 1 3 4 3 4 5 1000 1 2 2 4 1 3 4 3 1 4

Sample Output

3 0 1024 3213

Problem Source

ACM-ICPC Asia Thailand National On-Site Programming Contest 2015

题解：n的范围很小，所以可以用a*n+b来表示a,b所在位置的状态，用mp[i][j]表示a,b能否从i状态转移到j状态，再进行矩阵快速幂（矩阵的k次乘方就相当于i经过k条路到达j的方数 当k=2时，∑ way[i][x] * way[x][j]　(0<=x<n && x!=i && x!=j)，以此类推）

#include<iostream>
#include<algorithm> 
#include<string.h>
#include<stdio.h>
using namespace std;
typedef long long ll;
const int mod=9973;
const int maxn=105;
bool mp[105][105];
int t,n,m;
ll T;
struct matrix{
    int m[105][105];
    friend matrix operator*(matrix a,matrix b){
        matrix c;
        for(int i=0;i<n*n;i++)
           for(int j=0;j<n*n;j++){
                  c.m[i][j]=0;
                  for(int k=0;k<n*n;k++)
                  c.m[i][j]+=a.m[i][k]*b.m[k][j]%mod;
                  c.m[i][j]%=mod;
           }
        return c;
    }
}cnt;
matrix qpow(matrix a,ll b){
    matrix ret;
    memset(ret.m,0,sizeof(ret.m));
    for(int i=0;i<n;i++) ret.m[i][i]=1;
    while(b){
        if(b&1) ret=ret*a;
        b>>=1;
        a=a*a;
    }
    return ret;
}
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%I64d",&n,&m,&T);
        memset(mp,0,sizeof(mp));
        memset(cnt.m,0,sizeof(cnt.m));
        for(int u,v,i=0;i<m;i++){
            scanf("%d%d",&u,&v);
            mp[u][v]=mp[v][u]=1;
        } 
        for(int i=0;i<n*n;i++){
            int a=i/n+1,b=i%n+1;
            if(a==b) continue;
            for(int j=0;j<n*n;j++){
                int c=j/n+1,d=j%n+1;
                if(c==d||a==c||b==d||!mp[a][c]||!mp[b][d]) continue;
                cnt.m[i][j]=1;
            }
        }
        cnt=qpow(cnt,T-1);
        int ans=0;
        for(int i=0;i<n*n;i++){
            int a=i/n+1,b=i%n+1;
            if(!cnt.m[n-1][i]||a==b) continue;
            for(int j=1;j<=n;j++)
               if(mp[a][j]&&mp[b][j])
                  ans=(ans+cnt.m[n-1][i])%mod;
        }
        printf("%d\n",ans);
    }
}