[JSOI2008]Blue Mary的戰役地圖

標籤：cstring   col   ase   矩陣   tps   lap   namespace   har   last   

嘟嘟嘟

 

當看到n <= 50 的時候就樂呵了，暴力就行了，不過最暴力的方法是O(n7)……然後加一個二分邊長達到O(n6logn)，然後我們接著最佳化，把暴力比對改成O(1)的比對hash值，能達到O(n5logn),到勉強能過……不過我們還可以在最佳化一下，把第一個矩陣中所有邊長為 l 的子矩陣的hash值都存到一個數組中，然後sort一下，接著我們在枚舉第二個矩陣的子矩陣，然後在數組中用lower_bound的查詢就行。這樣的話複雜度應該是O(n3log(n2) * logn)了。

~~求一個矩陣的雜湊值就是每一行的雜湊值之和~~

 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;    //請無視 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 }

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[JSOI2008]Blue Mary的戰役地圖