BNUOJ 34990 Justice String

Justice String

Time Limit: 2000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

 

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

 

解题：Hash+二分

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 100100;
char sa[maxn],sb[maxn];
unsigned LL Ha[maxn],Hb[maxn],p = 131;
int len1,len2;
unsigned LL pp[maxn];
int calc(int a,int b){
    int high = min(len1 - a,len2 - b),low = 0,mid,ans = 0;
    while(low <= high){
        mid = (low + high)>>1;
        unsigned LL v1 = Ha[a] - Ha[a+mid]*pp[mid];
        unsigned LL v2 = Hb[b] - Hb[b+mid]*pp[mid];
        if(v1 == v2){
            ans = mid;
            low = mid + 1;
        }else high = mid - 1;
    }
    return ans;
}
int main() {
    int T,cs = 1;
    pp[0] = 1;
    for(int i = 1; i < maxn; ++i) pp[i] = pp[i-1]*p;
    scanf("%d",&T);
    while(T--){
        scanf("%s %s",sa,sb);
        len1 = strlen(sa);
        len2 = strlen(sb);
        Ha[len1] = 0;
        Hb[len2] = 0;
        for(int i = len1-1; i >= 0; --i) Ha[i] = Ha[i+1]*p + sa[i];
        for(int i = len2-1; i >= 0; --i) Hb[i] = Hb[i+1]*p + sb[i];
        int ans = -1;
        for(int i = 0; i <= len1 - len2; ++i){
            int sum = 0,a = i,b = 0,tmp = 0;
            tmp = calc(a,b);
            if(tmp >= len2-2){
                ans = i;
                break;
            }
            sum += tmp;
            a += tmp+1;
            b += tmp+1;
            tmp = calc(a,b);
            sum += tmp;
            if(sum >= len2-2){
                ans = i;
                break;
            }
            a += tmp+1;
            b += tmp+1;
            tmp = calc(a,b);
            sum += tmp;
            if(sum >= len2-2){
                ans = i;
                break;
            }
        }
        printf("Case #%d: %d
",cs++,ans);
    }
    return 0;
}