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

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 100100;

18 char sa[maxn],sb[maxn];

19 unsigned LL Ha[maxn],Hb[maxn],p = 131;

20 int len1,len2;

21 unsigned LL pp[maxn];

22 int calc(int a,int b){

23     int high = min(len1 - a,len2 - b),low = 0,mid,ans = 0;

24     while(low <= high){

25         mid = (low + high)>>1;

26         unsigned LL v1 = Ha[a] - Ha[a+mid]*pp[mid];

27         unsigned LL v2 = Hb[b] - Hb[b+mid]*pp[mid];

28         if(v1 == v2){

29             ans = mid;

30             low = mid + 1;

31         }else high = mid - 1;

32     }

33     return ans;

34 }

35 int main() {

36     int T,cs = 1;

37     pp[0] = 1;

38     for(int i = 1; i < maxn; ++i) pp[i] = pp[i-1]*p;

39     scanf("%d",&T);

40     while(T--){

41         scanf("%s %s",sa,sb);

42         len1 = strlen(sa);

43         len2 = strlen(sb);

44         Ha[len1] = 0;

45         Hb[len2] = 0;

46         for(int i = len1-1; i >= 0; --i) Ha[i] = Ha[i+1]*p + sa[i];

47         for(int i = len2-1; i >= 0; --i) Hb[i] = Hb[i+1]*p + sb[i];

48         int ans = -1;

49         for(int i = 0; i <= len1 - len2; ++i){

50             int sum = 0,a = i,b = 0,tmp = 0;

51             tmp = calc(a,b);

52             if(tmp >= len2-2){

53                 ans = i;

54                 break;

55             }

56             sum += tmp;

57             a += tmp+1;

58             b += tmp+1;

59             tmp = calc(a,b);

60             sum += tmp;

61             if(sum >= len2-2){

62                 ans = i;

63                 break;

64             }

65             a += tmp+1;

66             b += tmp+1;

67             tmp = calc(a,b);

68             sum += tmp;

69             if(sum >= len2-2){

70                 ans = i;

71                 break;

72             }

73         }

74         printf("Case #%d: %d\n",cs++,ans);

75     }

76     return 0;

77 }

View Code