标签：cstring col ase 矩阵 tps lap namespace har last
当看到n <= 50 的时候就乐呵了，暴力就行了，不过最暴力的方法是O(n7)……然后加一个二分边长达到O(n6logn)，然后我们接着优化，把暴力比对改成O(1)的比对hash值，能达到O(n5logn),到勉强能过……不过我们还可以在优化一下，把第一个矩阵中所有边长为 l 的子矩阵的hash值都存到一个数组中，然后sort一下，接着我们在枚举第二个矩阵的子矩阵，然后在数组中用lower_bound的查询就行。这样的话复杂度应该是O(n3log(n2) * logn)了。
~~求一个矩阵的哈希值就是每一行的哈希值之和~~
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(‘ ‘) 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 typedef long long ll; 16 typedef unsigned long long ull; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const int eps = 1e-8; 20 const int maxn = 55; 21 const ull base = 19260817; //请无视 22 inline ll read() 23 { 24 ll ans = 0; 25 char ch = getchar(), last = ‘ ‘; 26 while(!isdigit(ch)) {last = ch; ch = getchar();} 27 while(isdigit(ch)) {ans = ans * 10 + ch - ‘0‘; ch = getchar();} 28 if(last == ‘-‘) ans = -ans; 29 return ans; 30 } 31 inline void write(ll x) 32 { 33 if(x < 0) x = -x, putchar(‘-‘); 34 if(x >= 10) write(x / 10); 35 putchar(x % 10 + ‘0‘); 36 } 37 38 int n, a[2][maxn][maxn]; 39 ull has[2][maxn][maxn]; 40 ull f[maxn], b[maxn * maxn]; 41 int cnt = 0; 42 43 44 ull calc(int x, int y, int l, bool flag) 45 { 46 ull ret = 0; 47 for(int i = x; i <= x + l - 1; ++i) ret += has[flag][i][y + l - 1] - has[flag][i][y - 1] * f[l]; 48 return ret; 49 } 50 bool judge(int x) 51 { 52 cnt = 0; 53 for(int i = 1; i <= n - x + 1; ++i) 54 for(int j = 1; j <= n - x + 1; ++j) 55 b[++cnt] = calc(i, j, x, 0); 56 sort(b + 1, b + cnt + 1); 57 for(int i = 1; i <= n - x + 1; ++i) 58 for(int j = 1; j <= n - x + 1; ++j) 59 { 60 ull ha = calc(i, j, x, 1); 61 if(*lower_bound(b + 1, b +cnt + 1, ha) == ha) return 1; 62 } 63 return 0; 64 } 65 66 int main() 67 { 68 n = read(); 69 for(int k = 0; k <= 1; k++) 70 for(int i = 1; i <= n; ++i) 71 for(int j = 1; j <= n; ++j) a[k][i][j] = read(); 72 f[0] = 1; 73 for(int i = 1; i <= n; ++i) f[i] = f[i - 1] * base; 74 for(int k = 0; k <= 1; ++k) 75 for(int i = 1; i <= n; ++i) 76 for(int j = 1; j <= n; ++j) has[k][i][j] = has[k][i][j - 1] * base + a[k][i][j]; 77 int L = 1, R = n; 78 while(L + 1 < R) 79 { 80 int mid = (L + R) >> 1; 81 if(judge(mid)) L = mid; 82 else R = mid - 1; 83 } 84 write(judge(L + 1) ? L + 1 : L); enter; 85 return 0; 86 }
标签：cstring col ase 矩阵 tps lap namespace har last
原文地址：https://www.cnblogs.com/mrclr/p/9549230.html
