Luo Gu P3758 [TJOI2017] Coke

Luo Gu P3758 [TJOI2017] Coke

Description

• People Jiali Dun planet particularly fond of drinking cola. Thus, they developed a hostile planet a Coke robot and placed on the No. 1 city Jiali Dun planet. The cola robot has three acts: stay still, go to the next adjacent city, blew. It will be every second random trigger an action. Now give Jiali Dun planet city map, 0 seconds when Coke robot No. 1 in the city, and asked after t seconds, Coke robot behavior program number is how much?

Input

• The first line input condition two positive integers N, M, N represents the number of cities, M is the number of roads. (1 <= N <= 30,0 <M <100)

  The next M line input u, v, represents u, there is a road between v. (1 <= u, v <= n) to ensure that only one way between the two cities are connected.

  Finally, the input time t

Output

• Coke output of the robot's behavior program number, the answer may be large, please post the output of modulo 2017.

Sample Input

3 2
1 2
2 3
2

Sample Output

8

Data Size

• For 20% of the pn, there are 1 <t ≤ 1000

  To 100% of the pn, there are 1 <t ≤ 10 ^ 6.

answer:

• Fast Matrix power.
• Ideas from Zhang_RQ greatly

#include <iostream>
#include <cstdio>
#include <cstring>
#define N 35
#define mod 2017
using namespace std;

struct Node
{
    Node() {memset(m, 0, sizeof(m));}
    int m[N][N];
}a;
int n, m, k, ans;

Node cal(Node x, Node y)
{
    Node z;
    for(int i = 0; i <= n; i++)
        for(int j = 0; j <= n; j++)
            for(int k = 0; k <= n; k++)
                z.m[i][j] += ((x.m[i][k] % mod) * (y.m[k][j] % mod)) % mod,
                z.m[i][j] %= mod;
    return z;
}

Node power(Node a, int b)
{
    Node r = a, base = a;
    while(b)
    {
        if(b & 1) r = cal(r, base);
        base = cal(base, base);
        b >>= 1;
    }
    return r;
}

int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i++) a.m[i][0] = 1; //自爆
    for(int i = 0; i <= n; i++) a.m[i][i] = 1; //停留
    for(int i = 1; i <= m; i++)
    {
        int u, v; cin >> u >> v;
        a.m[u][v] = a.m[v][u] = 1;
    }
    cin >> k;
    a = power(a, k - 1);
    for(int i = 0; i <= n; i++)
        ans += a.m[1][i] % mod,
        ans %= mod;
    cout << ans;
    return 0;
}