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 = ;
 char sa[maxn],sb[maxn];
 unsigned LL Ha[maxn],Hb[maxn],p = ;
 int len1,len2;
 unsigned LL pp[maxn];
 int calc(int a,int b){
     int high = min(len1 - a,len2 - b),low = ,mid,ans = ;
     while(low <= high){
         mid = (low + high)>>;
         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 + ;
         }else high = mid - ;
     }
     return ans;
 }
 int main() {
     int T,cs = ;
     pp[] = ;
     for(int i = ; i < maxn; ++i) pp[i] = pp[i-]*p;
     scanf("%d",&T);
     while(T--){
         scanf("%s %s",sa,sb);
         len1 = strlen(sa);
         len2 = strlen(sb);
         Ha[len1] = ;
         Hb[len2] = ;
         for(int i = len1-; i >= ; --i) Ha[i] = Ha[i+]*p + sa[i];
         for(int i = len2-; i >= ; --i) Hb[i] = Hb[i+]*p + sb[i];
         int ans = -;
         for(int i = ; i <= len1 - len2; ++i){
             int sum = ,a = i,b = ,tmp = ;
             tmp = calc(a,b);
             if(tmp >= len2-){
                 ans = i;
                 break;
             }
             sum += tmp;
             a += tmp+;
             b += tmp+;
             tmp = calc(a,b);
             sum += tmp;
             if(sum >= len2-){
                 ans = i;
                 break;
             }
             a += tmp+;
             b += tmp+;
             tmp = calc(a,b);
             sum += tmp;
             if(sum >= len2-){
                 ans = i;
                 break;
             }
         }
         printf("Case #%d: %d\n",cs++,ans);
     }
     return ;
 }