hihocoder 1084 扩展KMP && 2014 北京邀请赛 Justice String

hihocoder 1084 : http://hihocoder.com/problemset/problem/1084

北京邀请赛 Just  String http://www.bnuoj.com/v3/problem_show.php?pid=34990

两道题同样的做法，题目基本内容是找到A的字串中和B串长度一样，且不同的字符个数不超过k个的置。

以hihocoder 1084为例， 是求有多少个A的字串的，与B串长度一样，且不同的字符个数不超过k。

分析：预处理hash，然后对每个字串进行判断，例如A[i~i+len-1] 和B[0~len-1], 首先找到一个A，B都相同的字符，然后从此出发找到长度尽可能大的完全相同的一截，这里可以用二分完成，然后再次找到一个对应位置A，B相同的字符，一旦不相同的字符个数超过k则立马终止，这样复杂度接近O(N*logN*k)。

代码：

 1 #include <bits/stdc++.h>
 2 #define pb push_back
 3 #define mp make_pair
 4 #define esp 1e-14
 5 #define lson   l, m, rt<<1
 6 #define rson   m+1, r, rt<<1|1
 7 #define sz(x) ((int)((x).size()))
 8 #define pf(x) ((x)*(x))
 9 #define pb push_back
10 #define pi acos(-1.0)
11 #define in freopen("solve_in.txt", "r", stdin);
12 #define bug(x) cerr << "Line : " << (x) <<  " >>>>>>\n";
13 #define TL cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC * 1000 << " ms" << endl;
14 #define inf 0x0f0f0f0f
15
16 using namespace std;
17 typedef long long LL;
18 typedef unsigned long long ULL;
19 typedef map<LL, int> MPS;
20 typedef pair<int, int> PII;
21 typedef MPS::iterator IT;
22
23 const int maxn = 100000 + 100;
24 const int B = (int)1e6 + 9;
25
26 char sa[maxn], sb[maxn];
27 ULL pw[maxn], ha[maxn], hb[maxn];
28 int n, m, k;
29 void pre(){
30     pw[0] = 1;
31     for(int i = 1; i < maxn; i++)
32         pw[i] = pw[i-1]*B;
33 }
34 inline ULL cal(ULL h[], int l, int r, int len){
35     return h[l]-h[r+1]*pw[len];
36 }
37 inline int idx(char ch){
38     return ch - ‘a‘;
39 }
40 int main() {
41
42     pre();
43     while(scanf("%s%s%d", sa, sb, &k) == 3){
44         n = strlen(sa);
45         m = strlen(sb);
46         ha[n] = hb[m] = 0;
47         for(int i = n-1; i >= 0; i--)
48             ha[i] = ha[i+1]*B+idx(sa[i]);
49         for(int i = m-1; i >= 0; i--)
50             hb[i] = hb[i+1]*B+idx(sb[i]);
51         int ans = 0;
52         for(int i = 0; i <= n-m; i++){
53             int j = 0, x = 0;
54 //            bug(i)
55             while(x <= k && j < m){
56                 while(j < m && sa[i+j] != sb[j] && x <= k)
57                     j++, x++;
58 //                bug(j)
59                 if(j >= m || x > k) break;
60                 int l = j, r = m;
61                 while(l+1 < r){
62                     int mid = (l+r)>>1;
63 //                    bug(mid)
64                     ULL h1 = cal(ha, i+l, i+mid, mid-l+1);
65                     ULL h2 = cal(hb, l, mid, mid-l+1);
66                     if(h1 == h2)
67                         l = mid;
68                     else r = mid;
69                 }
70                 if(l != m-1)
71                 x++;
72                 j = l+2;
73             }
74             ans += x <= k;
75         }
76         printf("%d\n", ans);
77     }
78     return 0;
79 }

北京邀请赛同样的分析方法。

代码：

 1 #include <bits/stdc++.h>
 2 #define pb push_back
 3 #define mp make_pair
 4 #define esp 1e-14
 5 #define lson   l, m, rt<<1
 6 #define rson   m+1, r, rt<<1|1
 7 #define sz(x) ((int)((x).size()))
 8 #define pf(x) ((x)*(x))
 9 #define pb push_back
10 #define pi acos(-1.0)
11 #define in freopen("solve_in.txt", "r", stdin);
12 #define bug(x) cerr << "Line : " << (x) <<  " >>>>>>\n";
13 #define TL cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC * 1000 << " ms" << endl;
14 #define inf 0x0f0f0f0f
15
16 using namespace std;
17 typedef long long LL;
18 typedef unsigned long long ULL;
19 typedef map<LL, int> MPS;
20 typedef pair<int, int> PII;
21 typedef MPS::iterator IT;
22
23 const int maxn = 100000 + 100;
24 const int B = (int)1e6 + 9;
25
26 char sa[maxn], sb[maxn];
27 ULL pw[maxn], ha[maxn], hb[maxn];
28 int n, m, k;
29 void pre() {
30     pw[0] = 1;
31     for(int i = 1; i < maxn; i++)
32         pw[i] = pw[i-1]*B;
33 }
34 inline ULL cal(ULL h[], int l, int r, int len) {
35     return h[l]-h[r+1]*pw[len];
36 }
37 inline int idx(char ch) {
38     return ch - ‘a‘;
39 }
40 int main() {
41
42     pre();
43     int T;
44     for(int t = scanf("%d", &T); t <= T; t++) {
45         k = 2;
46         scanf("%s%s", sa, sb);
47         n = strlen(sa);
48         m = strlen(sb);
49         ha[n] = hb[m] = 0;
50         for(int i = n-1; i >= 0; i--)
51             ha[i] = ha[i+1]*B+idx(sa[i]);
52         for(int i = m-1; i >= 0; i--)
53             hb[i] = hb[i+1]*B+idx(sb[i]);
54         int ans = -1;
55         for(int i = 0; i <= n-m; i++) {
56             int j = 0, x = 0;
57 //            bug(i)
58             while(x <= k && j < m) {
59                 while(j < m && sa[i+j] != sb[j] && x <= k)
60                     j++, x++;
61 //                bug(j)
62                 if(j >= m || x > k) break;
63                 int l = j, r = m;
64                 while(l+1 < r) {
65                     int mid = (l+r)>>1;
66 //                    bug(mid)
67                     ULL h1 = cal(ha, i+l, i+mid, mid-l+1);
68                     ULL h2 = cal(hb, l, mid, mid-l+1);
69                     if(h1 == h2)
70                         l = mid;
71                     else r = mid;
72                 }
73                 if(l != m-1)
74                     x++;
75                 j = l+2;
76             }
77             if(x <= k) {
78                 ans = i;
79                 break;
80             }
81         }
82         printf("Case #%d: %d\n", t, ans);
83     }
84     return 0;
85 }