2 条题解
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1
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 110; ll T; int n, m, lim, len[N], act, now[10][10]; char s[10][10], t[10][10]; struct mat { ll a[65][65]; mat() {memset(a, 0, sizeof(a));} mat operator * (mat &x) { mat ans; for(int i = 0; i <= lim; ++i) for(int j = 0; j <= lim; ++j) for(int k = 0; k <= lim; ++k) ans.a[i][j] += a[i][k] * x.a[k][j]; return ans; } mat Pow(ll t, mat a) { mat ans; ans.a[0][0] = 1; while(t) { if(t & 1) ans = ans * a; a = a * a; t >>= 1; } return ans; } }c[61], C; int idx(int x, int y) {return (x - 1) * m + y;} void print(mat x, bool flag) { puts("#####################"); if(flag) { for(int i = 0; i <= lim; ++i) printf("%lld ", x.a[0][i]); puts(""); return; } for(int i = 0; i <= lim; ++i) { printf("%d: ", i); for(int j = 0; j <= lim; ++j) { printf("%lld ", x.a[i][j]); } puts(""); } puts("#####################"); } int main() { scanf("%d%d%lld%d", &n, &m, &T, &act); lim = n * m; for(int i = 1; i <= n; ++i) { scanf("%s", t[i] + 1); for(int j = 1; j <= m; ++j) t[i][j] -= '0', ++t[i][j]; } for(int i = 1; i <= act; ++i) { scanf("%s", s[i] + 1); len[i] = strlen(s[i] + 1); } for(int pos = 1; pos <= 60; ++pos) { c[pos].a[0][0] = 1; for(int i = 1; i <= n; ++i) { for(int j = 1; j <= m; ++j) { // putchar(s[t[i][j]][now[i][j]]); now[i][j] = (now[i][j] % len[t[i][j]]) + 1; if(s[t[i][j]][now[i][j]] >= '0' && s[t[i][j]][now[i][j]] <= '9') { c[pos].a[0][idx(i, j)] = s[t[i][j]][now[i][j]] - '0'; c[pos].a[idx(i, j)][idx(i, j)] = 1; } else { if(s[t[i][j]][now[i][j]] == 'D') c[pos].a[idx(i, j)][idx(i, j)] = 0; if(i > 1 && s[t[i][j]][now[i][j]] == 'N') c[pos].a[idx(i, j)][idx(i - 1, j)] = 1; if(i < n && s[t[i][j]][now[i][j]] == 'S') c[pos].a[idx(i, j)][idx(i + 1, j)] = 1; if(j > 1 && s[t[i][j]][now[i][j]] == 'W') c[pos].a[idx(i, j)][idx(i, j - 1)] = 1; if(j < m && s[t[i][j]][now[i][j]] == 'E') c[pos].a[idx(i, j)][idx(i, j + 1)] = 1; } } } // puts(""); } C = c[1]; for(int pos = 2; pos <= 60; ++pos) C = C * c[pos]; C = C.Pow(T / 60, C); for(int pos = 1; pos <= T % 60; ++pos) { // printf("Round # %d\n", pos); C = C * c[pos]; // print(c[pos], 0); // printf("The test # %d\n", pos); // print(C, 1); } ll ans = 0; for(int i = 1; i <= lim; ++i) ans = max(ans, C.a[0][i]); printf("%lld\n", ans); return 0; }
信息
- ID
- 117
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 5
- 标签
- 递交数
- 25
- 已通过
- 12
- 上传者