2 条题解
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1问号 (周文浩) LV 10 @ 2024-5-13 17:05:52
#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; }
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12023-9-29 14:53:47@
#include <bits/stdc++.h> #include <unordered_map>
using namespace std; #define ENDL "\n" #define lowbit(x) (x & (-x)) typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair<int, int> pii; typedef pair<ll, ll> pll; typedef pair<double, double> pdd; const double eps = 1e-8; const double pi = acos(-1.0); const int inf = 0x3f3f3f3f; const double dinf = 1e300; const ll INF = 1e18; const int Mod = 1e9 + 7; const int maxn = 2e5 + 10;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
struct Matrix { int n, m; ll a[65][65];
Matrix() {} Matrix(int x, int y) { n = x, m = y; memset(a, 0, sizeof a); } Matrix operator*(const Matrix &p) const { Matrix ret(n, p.m); for (int i = 0; i <= n; i++) for (int j = 0; j <= p.m; j++) for (int k = 0; k <= m; k++) ret.a[i][j] += a[i][k] * p.a[k][j]; return ret; } void print() { cout << "-----------------" << endl; for (int i = 0; i <= n; i++) { for (int j = 0; j <= m; j++) cout << a[i][j] << " "; cout << endl; } }
} t[65];
Matrix qkp(Matrix mx, int q) { Matrix ans(mx.n, mx.m); for (int i = 0; i <= ans.n; i++) ans.a[i][i] = 1; while (q) { if (q & 1) ans = ans * mx; mx = mx * mx; q >>= 1; } return ans; }
string s[15]; int b[10][10]; int n, m, num, cnt, tot;
inline int ID(int p, int q) { return n * (p - 1) + q; }
int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0); cin >> n >> m >> num >> cnt; for (int i = 1; i <= n; i++) { cin >> s[0]; for (int j = 1; j <= s[0].size(); j++) { b[i][j] = s[0][j - 1] - '0'; } } for (int i = 0; i < cnt; i++) cin >> s[i]; tot = 1; for (int i = 0; i < cnt; i++) tot = lcm(tot, (int) s[i].size()); for (int i = 0; i < cnt; i++) { int x = tot / (int) s[i].size(); string res = ""; for (int j = 1; j <= x; j++) res += s[i]; s[i] = res; } Matrix G(n * m, n * m); for (int i = 0; i <= n * m; i++) G.a[i][i] = 1; for (int k = 1; k <= tot; k++) { //求出每秒的转移矩阵 t[k].n = t[k].m = n * m, t[k].a[0][0] = 1; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) { char ch = s[b[i][j]][k - 1]; if (ch >= '0' && ch <= '9') t[k].a[0][ID(i, j)] = ch - '0', t[k].a[ID(i, j)][ID(i, j)] = 1; else if (ch == 'N' && i > 1) t[k].a[ID(i, j)][ID(i - 1, j)] = 1; else if (ch == 'S' && i < n) t[k].a[ID(i, j)][ID(i + 1, j)] = 1; else if (ch == 'W' && j > 1) t[k].a[ID(i, j)][ID(i, j - 1)] = 1; else if (ch == 'E' && j < m) t[k].a[ID(i, j)][ID(i, j + 1)] = 1; } // t[k].print(); G = G * t[k]; } int x = num / tot, y = num % tot; Matrix F(0, n * m); F.a[0][0] = 1; G = qkp(G, x); F = F * G; for (int k = 1; k <= y; k++) F = F * t[k]; ll ans = 0; for (int i = 1; i <= n * m; i++) ans = max(ans, F.a[0][i]); cout << ans << ENDL; return 0; }
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信息
- ID
- 117
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 6
- 标签
- 递交数
- 24
- 已通过
- 11
- 上传者