説明
解決
以降\（GCD（1,2,3,4,5,6）60 \ =）を、そのようにそれぞれ（\ 60秒）の\固定することができる内変換（\ 60 \）行列、\は（ \ FORALL X \ \）[1,60]で、そう[0] = [0]（[X] \ 1 \）の安定したストリームの一定の源として
\（F [I] \）からのラインである\ （0 \）する（0 \）\、からカラム\（0 \）に\（N \ AST Mが\）を行列表し\（Iは\）状態秒、\（F. [0] [0を] = 1 \）を
行う\（T = 60 * P + Q \） 、そう\（A \）の\（60 \）は行列が乗算され、\（F. [0] \ P * AST A ^ \ ^のProd {Q} _ {i = 1 } F [i]が\） 最終的なマトリックスであっ
に行列内の最大値を見つけます
コード
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define num(x, y) ((x - 1) * m + y)
#define MAXN 110
struct Mat {
int x, y;
int val[MAXN][MAXN];
} f[61];
Mat operator * (const Mat &a, const Mat &b) {
Mat c;
c.x = a.x;
c.y = b.y;
for (register int i = 0; i <= c.x; i++)
for (register int j = 0; j <= c.y; j++) {
c.val[i][j] = 0;
for (register int k = 0; k <= a.y; k++)
c.val[i][j] += a.val[i][k] * b.val[k][j];
}
return c;
}
int n, m, t, act, X, Y, ans;
int c[MAXN], Len[MAXN];
char str[MAXN], s[MAXN][MAXN];
Mat a, HSH;
inline int read() {
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
Mat Pow(Mat a, int k) {
Mat ans;
while (k) {
if (k % 2) ans = ans * a;
a = a * a;
k /= 2;
}
return ans;
}
int main() {
n = read(), m = read(), t = read(), act = read();
for (register int i = 1; i <= n; i++) {
scanf("%s", str + 1);
for (register int j = 1; j <= m; j++)
c[num(i, j)] = (str[j] ^ 48);
}
for (register int i = 0; i < act; i++) {
scanf("%s", s[i] + 1);
Len[i] = strlen(s[i] + 1);
}
a.x = 0, a.y = 64;
a.val[0][0] = 1;
for (register int k = 1; k <= 60; k++) {
f[k].x = 60;
f[k].y = 60;
f[k].val[0][0] = 1;
for (register int i = 1; i <= n; i++)
for (register int j = 1; j <= m; j++) {
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == '1') f[k].val[0][num(i, j)] = 1;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == '2') f[k].val[0][num(i, j)] = 2;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == '3') f[k].val[0][num(i, j)] = 3;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == '4') f[k].val[0][num(i, j)] = 4;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == '5') f[k].val[0][num(i, j)] = 5;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == '6') f[k].val[0][num(i, j)] = 6;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == '7') f[k].val[0][num(i, j)] = 7;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == '8') f[k].val[0][num(i, j)] = 8;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == '9') f[k].val[0][num(i, j)] = 9;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == 'N') f[k].val[num(i, j)][num(i - 1, j)] = 1;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == 'W') f[k].val[num(i, j)][num(i, j - 1)] = 1;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == 'S') f[k].val[num(i, j)][num(i + 1, j)] = 1;
if (s[c[num(i, j)]][(k - 1) % (Len[c[num(i, j)]]) + 1] == 'E') f[k].val[num(i, j)][num(i, j + 1)] = 1;
}
}
for (register int i = 0; i <= 64; i++)
HSH.val[i][i] = 1;
for (register int i = 1; i <= 60; i++)
HSH = HSH * f[i];
X = t / 60;
Y = t % 60;
if (X) a = a * Pow(HSH, X);
for (register int i = 1; i <= Y; i++)
a = a * f[i];
for (register int i = 1; i <= 64; i++)
ans = max(ans, a.val[0][i]);
printf("%d", ans);
return 0;
}