[JSOI2008]Blue Mary的战役地图

嘟嘟嘟

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 }

原文地址：https://www.cnblogs.com/mrclr/p/9549230.html