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NEUQ 2015: Bitmap(二维hash)
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題目鏈接
題意
給一個N×NN × NN×N的矩陣問包含多少個M×MM×MM×M的子矩陣，子矩陣不一定完全相同，同時加上某個數相同也算
思路
首先差分，這樣就可以直接找匹配的矩陣。
- 二維hash+容斥判斷矩陣是否相同
- 沖突是不可避免的，我們要最小化沖突，可以對矩陣進行左右和上下兩次差分，進行計算
- 可以用unsigned long long，也可以取模。
- 一直wa的原因是：差分之后矩陣的大小搞錯了，矩陣原來是N×NN×NN×N差分之后應該是N×N?1N×N-1N×N?1或者N?1×NN-1 × NN?1×N
#include <bits/stdc++.h>
#define LL long long
#define uLL long long
#define P pair<int, int>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
using namespace std
;
const int maxn
= 2003;
LL a
[maxn
][maxn
], b
[maxn
][maxn
], c
[maxn
][maxn
], d
[maxn
][maxn
];
LL pow1
[maxn
], pow2
[maxn
];
int t1
[maxn
][maxn
], t2
[maxn
][maxn
];
const LL seed1
= 805306457;
const LL seed2
= 402653189;
const LL mod
= 1e9 + 7;int n
, m
;
void init() {pow2
[0] = pow1
[0] = 1;for (int i
= 1; i
<= n
; ++i
) {pow1
[i
] = pow1
[i
-1] * seed1
% mod
;pow2
[i
] = pow2
[i
-1] * seed2
% mod
;}
}
void solve(LL x
[][maxn
], int y
) {for (int i
= 1; i
<= y
; ++i
) {for (int j
= 1; j
<= y
; ++j
) {x
[i
][j
] = (x
[i
][j
] + x
[i
][j
-1] * seed1
) % mod
;}for (int j
= 1; j
<= y
; ++j
) {x
[i
][j
] = (x
[i
][j
] + x
[i
-1][j
] * seed2
) % mod
;}}
}
void print(LL x
[][maxn
], int l
, int r
) {for (int i
= 1; i
<= l
; ++i
) {for (int j
= 1; j
<= r
; ++j
) {cout
<< x
[i
][j
] << " ";}cout
<< endl
;}
}
int sum
[maxn
][maxn
];
int main
() {ios
::sync_with_stdio(0);cin
.tie(0), cout
.tie(0);cin
>> n
>> m
;for (int i
= 1; i
<= n
; ++i
) {for (int j
= 1; j
<= n
; ++j
) {cin
>> t1
[i
][j
];}}for (int i
= 1; i
<= n
; ++i
) {for (int j
= 1; j
<= n
; ++j
) {a
[i
][j
] = (t1
[i
][j
+1] - t1
[i
][j
] + mod
) % mod
;b
[j
][i
] = (t1
[j
+1][i
] - t1
[j
][i
] + mod
) % mod
;}}for (int i
= 1; i
<= m
; ++i
) {for (int j
= 1; j
<= m
; ++j
) {cin
>> t2
[i
][j
];}}for (int i
= 1; i
<= m
; ++i
) {for (int j
= 1; j
<= m
; ++j
) {c
[i
][j
] = (t2
[i
][j
+1] - t2
[i
][j
] + mod
) % mod
;d
[j
][i
] = (t2
[j
+1][i
] - t2
[j
][i
] + mod
) % mod
;}}init();solve(a
, n
);solve(b
, n
);solve(c
, m
);solve(d
, m
); for (int i
= m
; i
<= n
; ++i
) {for (int j
= m
-1; j
<= n
-1; ++j
) {LL tmp
= a
[i
][j
];tmp
= (tmp
- a
[i
-m
][j
] * pow2
[m
] % mod
+ mod
) % mod
;tmp
= (tmp
- a
[i
][j
-m
+1] * pow1
[m
-1] % mod
+ mod
) % mod
;tmp
= (tmp
+ (a
[i
-m
][j
-m
+1] * pow1
[m
-1] % mod
) * pow2
[m
] % mod
) % mod
;if (tmp
== c
[m
][m
-1]) sum
[i
-m
][j
-m
+1]++;}}for (int i
= m
-1; i
<= n
-1; ++i
) {for (int j
= m
; j
<= n
; ++j
) {LL tmp
= b
[i
][j
];tmp
= (tmp
- b
[i
-m
+1][j
] * pow2
[m
-1] % mod
+ mod
) % mod
;tmp
= (tmp
- b
[i
][j
-m
] * pow1
[m
] % mod
+ mod
) % mod
;tmp
= (tmp
+ (b
[i
-m
+1][j
-m
] * pow1
[m
] % mod
) * pow2
[m
-1] % mod
) % mod
;if (tmp
== d
[m
-1][m
]) sum
[i
-m
+1][j
-m
]++;}}int ans
= 0;for (int i
= 0; i
<= n
; ++i
) {for (int j
= 0; j
<= n
; ++j
) {if (sum
[i
][j
] == 2) ans
++;}}cout
<< ans
<< endl
;return 0;
}
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