Flea Circus（Project Euler 213）

original version

hackerrank programming version

题目大意是N*N的格子，每个格子一开始有1个跳蚤，每过单位时间跳蚤会等概率向四周跳，问M秒后空格子的期望个数。

题解：

对于每个跳蚤暴力模拟每一秒，算出M秒后它到各个格子的概率最后统计就好了，hackerrank上的版本需要常数优化，比如根据对称性只考虑左上四分之一块矩形.

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int mod = 1e9 + 7;
#define MAXN 42
#define MP make_pair
const int INF = 1e9 + 10;

int n, m;
int s[MAXN][MAXN], tmp[MAXN][MAXN], cnt[MAXN][MAXN];
LL p[MAXN][MAXN];
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};
int inv[5];


int power(int x, int p)
{
    int res = 1;
    for (; p; p >>= 1)
    {
        if (p & 1) res = 1LL * res * x % mod;
        x = 1LL * x * x % mod;
    }
    return res;
}

void work(int sx, int sy)
{
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j)
            s[i][j] = 0;
    s[sx][sy] = 1;
     
    int x, y;
    for (int _m = 1; _m <= m; ++_m)
    {
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                tmp[i][j] = 0;
        for (int i = 1; i <= n; ++i)
        {
            for (int j = 1; j <= n; ++j)
            {
                for (int d = 0; d < 4; ++d)
                {
                    x = i + dx[d];
                    y = j + dy[d];
                    (tmp[x][y] += 1LL * cnt[i][j] * s[i][j] % mod) %= mod;
                }
            }
        }
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                s[i][j] = tmp[i][j];
    }
    int t;
    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j <= n; ++j)
        {
            t = 1 - s[i][j] + mod;
            (p[i][j] *= t) %= mod;
            (p[n + 1 - i][j] *= t) %= mod;
            (p[i][n + 1 - j] *= t) %= mod;
            (p[n + 1 - i][n + 1 - j] *= t) %= mod;
        }
    }
}



int main() 
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);

    inv[1] = 1;
    inv[2] = power(2, mod - 2);
    inv[3] = power(3, mod - 2);
    inv[4] = power(4, mod - 2);
    
    int T, x, y;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d %d", &n, &m);
        for (int i = 1; i <= n; ++i)
        {
            for (int j = 1; j <= n; ++j)
            {
                cnt[i][j] = 0;
                for (int d = 0; d < 4; ++d)
                {
                    x = i + dx[d];
                    y = j + dy[d];
                    if (x >= 1 && x <= n && y >= 1 && y <= n)
                        ++cnt[i][j];
                }
                cnt[i][j] = inv[cnt[i][j]];
            }
            
        }
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                p[i][j] = 1;
        LL ans = 0;
        for (int i = 1; i <= n / 2; ++i)
            for (int j = 1; j <= n / 2; ++j)
                work(i, j);
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                ans += p[i][j];
        ans %= mod;
        cout << ans << endl;
     }
    
    return 0; 
}