BNUOJ 34990 Justice String 字符串hash

Justice String

Given two strings A and B, your task is to find a substring of A called justice string, which has the same length as B, and only has at most two characters different from B.

Input

The first line of the input contains a single integer T, which is the number of test cases.

For each test case, the first line is string A, and the second is string B.

Both string A and B contain lowercase English letters from  a to  z only. And the length of these two strings is between 1 and 100000, inclusive. 

 

Output

For each case, first output the case number as " Case #x: ", and x is the case number. Then output a number indicating the start position of substring C in A, position is counted from 0. If there is no such substring C, output -1.

And if there are multiple solutions, output the smallest one. 

 

Sample Input

3
aaabcd
abee
aaaaaa
aaaaa
aaaaaa
aabbb

Sample Output

Case #1: 2
Case #2: 0
Case #3: -1

Source

2014 ACM-ICPC Beijing Invitational Programming Contest

Tags ( Click to see )

被称做sb hash的题目。。 

代码很短。就是每次从两个位置开始找公共前缀。 找到后，justice用一次，接着找。知道justice用完。。从第一个位置开始判断。 二分查找思想实现求公共前缀。。

#include <cstdio>
#include <cstdlib>

const int N = 111111;
#define LL long long
const long long MOD = 1e9+7;
const int SEED = 31;
LL base[N];
char s1[N],s2[N];

LL a[N],b[N];
int n1,n2;
int len;
void format(char s[],LL rec[],int &len){
	LL ret = 0;
	int i;
	for(i = 1; s[i] ;i++){
		ret = (ret * SEED + s[i]) % MOD;
		rec[i] = ret;
	}
	len = i-1;
}
LL calc(LL F[],int l,int r){
	return ((F[r]-F[l-1]*base[r-l+1])%MOD+MOD)%MOD;
}
int cl(int i,int j){
	int ret = 0;
	int l = 1,r = n2-j+1;
	while(l <= r){
		int mid = (l+r)>>1;
		if(calc(a,i,i+mid-1) == calc(b,j,j+mid-1) ){
			l = mid + 1;
			ret = mid ;
		}else r = mid-1;
	}
	return ret;
}
bool check(int start){
	int use = 0;
	int i,j=1;
	for(i = start; i <= n1 && use < 2 && j < n2; i++,j++){
		int logest = cl(i,j);
		i += logest ;
		j += logest ;
		use++;
		if(use >= 2 && j < n2){
			logest = cl(i+1,j+1);
			j += logest;
			if(j >= n2)return true;
			else return false;
		}
	}
	return true; 
}
int main(){
	int t,tt=0;
	scanf("%d",&t);
	base[0] = 1;
	for(int i = 1;i < N;i++){
		base[i] = base[i-1] * SEED % MOD;
	}
	while(t--){
		scanf("%s %s",s1+1,s2+1);
		format(s1,a,n1);
		format(s2,b,n2);
		int ret = -1;
		for(int i = 1;i <= n1 - n2 + 1;i++){
			if(check(i)){
				ret = i-1;
				break;
			}
		}
		printf("Case #%d: %d\n",++tt,ret);
	}
	return 0;
}