UVALive - 6893 The Big Painting 字符串哈希

题目链接：

http://acm.hust.edu.cn/vjudge/problem/129730

The Big Painting

Time Limit: 5000MS

题意

给你一个模板串和待匹配串，问模板串在待匹配串中出现的次数（这里的串是二维矩阵）

题解

每一行做前缀和哈希。
统计的时候先按列再按行，这样在做行的话我们可以利用滚动的形式，计算纵向的哈希值（既总的哈希值）

代码

include

include

include

include

include

include

include

include

include

using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) ;//cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)

typedef long long LL;
typedef vector VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;

//start----------------------------------------------------------------------

const int maxn=2020;
//P进制哈希
const int P=123;

unsigned long long H[maxn][maxn],xp[maxn*maxn],Val;

int n1,m1,n2,m2;

char s1[maxn][maxn],s2[maxn][maxn];
//预处理出位权
void pre() {
xp[0]=1;
rep(i,1,maxnmaxn) xp[i]=xp[i-1]P;
}
//处理前缀和的哈希值
void get_H() {
Val=0;
rep(i,1,n1+1) rep(j,1,m1+1) {
Val=Val*P+(s1[i][j]-'a'+1);
}

clr(H,0);
rep(i,1,n2+1) rep(j,1,m2+1) {
H[i][j]=H[i][j-1]P+(s2[i][j]-'a'+1);
}
}
//枚举，匹配
void solve() {
int ans=0;
for(int j=1; j<=m2-m1+1; j++) {
unsigned long long tmp=0,tmp2=0;
int lj=j-1,rj=j+m1-1;
int i;
for(i=1;i<=n1;i++){
tmp=tmpxp[m1]+H[i][rj]-H[i][lj]xp[m1];
}
do{
if(Val==tmp) ans++;
tmp-=(H[i-n1][rj]-H[i-n1][lj]xp[m1])xp[m1(n1-1)];
tmp=tmpxp[m1]+H[i][rj]-H[i][lj]xp[m1];
i++;
}while(i<=n2+1);
}
printf("%d\n",ans);
}

int main() {
pre();
while(scanf("%d%d%d%d",&n1,&m1,&n2,&m2)==4&&n1) {
rep(i,1,n1+1) scanf("%s",s1[i]+1);
rep(i,1,n2+1) scanf("%s",s2[i]+1);
get_H();
solve();
}
return 0;
}

//end-----------------------------------------------------------------------

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