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-8
# define MOD 100000000
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int N=1005;
//Code begin...

char s[N][N], str[N][N];
int hah[N][N], has[N][N], M1[N], M2[N], key1=1789, key2=131;
VI v;

int main ()
{
    int n, m, a, b, K, x, y;
    M1[0]=M2[0]=1; FO(i,1,N) M1[i]=M1[i-1]*key1, M2[i]=M2[i-1]*key2;
    scanf("%d%d%d%d",&n,&m,&a,&b);
    FOR(i,1,n) scanf("%s",s[i]+1);
    FOR(i,1,n) FOR(j,1,m) hah[i][j]=hah[i][j-1]*key1+s[i][j];
    FOR(i,1,n) FOR(j,1,m) hah[i][j]+=hah[i-1][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,1,a) scanf("%s",str[i]+1);
        FOR(i,1,a) FOR(j,1,b) has[i][j]=has[i][j-1]*key1+str[i][j];
        FOR(i,1,a) FOR(j,1,b) has[i][j]+=has[i-1][j]*key2;
        y=lower_bound(v.begin(),v.end(),has[a][b])-v.begin();
        puts(y<v.size()&&v[y]==has[a][b]?"1":"0");
    }
    return 0;
}
View Code

 

posted @ 2017-06-02 13:28  free-loop  阅读(...)  评论(...编辑  收藏