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 }