POJ 3690 Constellations hash

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题意：给出一个N*M的大矩阵。在给出T个P*Q的小矩阵，问有多少个小矩阵在大矩阵中出现过。

思路：二维hash。算出整个矩阵的所有的P*Q小矩阵的hash值，同时求出目标矩阵的hash值，判断该值是否出现即可。

复杂度：N * M * K

代码如下：

#include <cstdio>#include <algorithm>#include <cstring>#include <set>using namespace std;typedef unsigned long long  ull;const int MAX_SIZE = 1010;const int MAX_T = 110;int N,M,T,P,Q;char field[MAX_SIZE][MAX_SIZE];char patterns[MAX_T][MAX_SIZE][MAX_SIZE];ull has[MAX_SIZE][MAX_SIZE],tmp[MAX_SIZE][MAX_SIZE];int cas;void compute_hash(char a[MAX_SIZE][MAX_SIZE],int n, int m){    const ull B1 = 9973;    const ull B2 = 100000007;    ull t1 = 1;    for(int i = 0; i < Q; ++i) t1 *= B1;    for(int i = 0; i < n; ++i){        ull e = 0;        for(int j = 0; j < Q; ++j) e = e * B1 + a[i][j];        for(int j = 0; j + Q <= m; ++j){            tmp[i][j] = e;            if(j + Q < m)                e = e * B1 - t1 * a[i][j] + a[i][j + Q];        }    }    ull t2 = 1;    for(int i = 0; i < P; ++i) t2 *= B2;    for(int j = 0; j + Q <= m; ++j){        ull e = 0;        for(int i = 0; i < P; ++i) e = e * B2 + tmp[i][j];        for(int i = 0; i + P <= n; ++i){            has[i][j] = e;            if(i + P < n) e = e * B2 - t2 * tmp[i][j] + tmp[i+P][j];        }    }}void solve(){    multiset<ull> unseen;    compute_hash(field,N,M);    for(int i = 0; i + P <= N; ++i)        for(int j = 0; j + Q <= M; ++j)            unseen.insert(has[i][j]);    int cnt = 0;    for(int k = 0; k < T; ++k){        compute_hash(patterns[k],P,Q);        if(unseen.count(has[0][0])) cnt++;    }    printf("Case %d: %d\n",++cas ,cnt);}int main(void){    //freopen("input.txt","r",stdin);    while(scanf("%d%d%d%d%d",&N,&M,&T,&P,&Q),N||M||T||P||Q){        for(int i = 0; i < N; ++i)            scanf("%s",field[i]);        for(int i = 0; i < T; ++i){            for(int j = 0; j < P; ++j)                scanf("%s",patterns[i][j]);        }        solve();    }    return 0;}

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