BZOJ 2462 矩阵模板(二维hash)

题意:给出一个n*m的01矩阵，以及k个a*b的01矩阵，问每个是否能匹配原来的01矩阵。

由于k个矩阵的长和宽都是一样的，所以把原矩阵的所有a*b的子矩阵给hash出来。然后依次查找是否存在即可。

map被卡，用lower_bound即可。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <bitset>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

const int N=;

//Code begin...

char s[N][N], str[N][N];

int hah[N][N], has[N][N], M1[N], M2[N], key1=, key2=;

VI v;

int main ()

{

    int n, m, a, b, K, x, y;

    M1[]=M2[]=; FO(i,,N) M1[i]=M1[i-]*key1, M2[i]=M2[i-]*key2;

    scanf("%d%d%d%d",&n,&m,&a,&b);

    FOR(i,,n) scanf("%s",s[i]+);

    FOR(i,,n) FOR(j,,m) hah[i][j]=hah[i][j-]*key1+s[i][j];

    FOR(i,,n) FOR(j,,m) hah[i][j]+=hah[i-][j]*key2;

    FOR(i,a,n) FOR(j,b,m) {

        x=hah[i][j]-hah[i-a][j]*M2[a]-hah[i][j-b]*M1[b]+hah[i-a][j-b]*M2[a]*M1[b];

        v.pb(x);

    }

    sort(v.begin(),v.end());

    scanf("%d",&K);

    while (K--) {

        FOR(i,,a) scanf("%s",str[i]+);

        FOR(i,,a) FOR(j,,b) has[i][j]=has[i][j-]*key1+str[i][j];

        FOR(i,,a) FOR(j,,b) has[i][j]+=has[i-][j]*key2;

        y=lower_bound(v.begin(),v.end(),has[a][b])-v.begin();

        puts(y<v.size()&&v[y]==has[a][b]?"":"");

    }

    return ;

}