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：

 #include <bits/stdc++.h>
 #define pb push_back
 #define mp make_pair
 #define esp 1e-14
 #define lson   l, m, rt<<1
 #define rson   m+1, r, rt<<1|1
 #define sz(x) ((int)((x).size()))
 #define pf(x) ((x)*(x))
 #define pb push_back
 #define pi acos(-1.0)
 #define in freopen("solve_in.txt", "r", stdin);
 #define bug(x) cerr << "Line : " << (x) <<  " >>>>>>\n";
 #define TL cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC * 1000 << " ms" << endl;
 #define inf 0x0f0f0f0f

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 typedef map<LL, int> MPS;
 typedef pair<int, int> PII;
 typedef MPS::iterator IT;

 const int maxn =  + ;
 const int B = (int)1e6 + ;

 char sa[maxn], sb[maxn];
 ULL pw[maxn], ha[maxn], hb[maxn];
 int n, m, k;
 void pre(){
     pw[] = ;
     for(int i = ; i < maxn; i++)
         pw[i] = pw[i-]*B;
 }
 inline ULL cal(ULL h[], int l, int r, int len){
     return h[l]-h[r+]*pw[len];
 }
 inline int idx(char ch){
     return ch - 'a';
 }
 int main() {

     pre();
     while(scanf("%s%s%d", sa, sb, &k) == ){
         n = strlen(sa);
         m = strlen(sb);
         ha[n] = hb[m] = ;
         for(int i = n-; i >= ; i--)
             ha[i] = ha[i+]*B+idx(sa[i]);
         for(int i = m-; i >= ; i--)
             hb[i] = hb[i+]*B+idx(sb[i]);
         int ans = ;
         for(int i = ; i <= n-m; i++){
             int j = , x = ;
 //            bug(i)
             while(x <= k && j < m){
                 while(j < m && sa[i+j] != sb[j] && x <= k)
                     j++, x++;
 //                bug(j)
                 if(j >= m || x > k) break;
                 int l = j, r = m;
                 while(l+ < r){
                     int mid = (l+r)>>;
 //                    bug(mid)
                     ULL h1 = cal(ha, i+l, i+mid, mid-l+);
                     ULL h2 = cal(hb, l, mid, mid-l+);
                     if(h1 == h2)
                         l = mid;
                     else r = mid;
                 }
                 if(l != m-)
                 x++;
                 j = l+;
             }
             ans += x <= k;
         }
         printf("%d\n", ans);
     }
     return ;
 }

北京邀请赛同样的分析方法。

代码：

 #include <bits/stdc++.h>
 #define pb push_back
 #define mp make_pair
 #define esp 1e-14
 #define lson   l, m, rt<<1
 #define rson   m+1, r, rt<<1|1
 #define sz(x) ((int)((x).size()))
 #define pf(x) ((x)*(x))
 #define pb push_back
 #define pi acos(-1.0)
 #define in freopen("solve_in.txt", "r", stdin);
 #define bug(x) cerr << "Line : " << (x) <<  " >>>>>>\n";
 #define TL cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC * 1000 << " ms" << endl;
 #define inf 0x0f0f0f0f

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 typedef map<LL, int> MPS;
 typedef pair<int, int> PII;
 typedef MPS::iterator IT;

 const int maxn =  + ;
 const int B = (int)1e6 + ;

 char sa[maxn], sb[maxn];
 ULL pw[maxn], ha[maxn], hb[maxn];
 int n, m, k;
 void pre() {
     pw[] = ;
     for(int i = ; i < maxn; i++)
         pw[i] = pw[i-]*B;
 }
 inline ULL cal(ULL h[], int l, int r, int len) {
     return h[l]-h[r+]*pw[len];
 }
 inline int idx(char ch) {
     return ch - 'a';
 }
 int main() {

     pre();
     int T;
     for(int t = scanf("%d", &T); t <= T; t++) {
         k = ;
         scanf("%s%s", sa, sb);
         n = strlen(sa);
         m = strlen(sb);
         ha[n] = hb[m] = ;
         for(int i = n-; i >= ; i--)
             ha[i] = ha[i+]*B+idx(sa[i]);
         for(int i = m-; i >= ; i--)
             hb[i] = hb[i+]*B+idx(sb[i]);
         int ans = -;
         for(int i = ; i <= n-m; i++) {
             int j = , x = ;
 //            bug(i)
             while(x <= k && j < m) {
                 while(j < m && sa[i+j] != sb[j] && x <= k)
                     j++, x++;
 //                bug(j)
                 if(j >= m || x > k) break;
                 int l = j, r = m;
                 while(l+ < r) {
                     int mid = (l+r)>>;
 //                    bug(mid)
                     ULL h1 = cal(ha, i+l, i+mid, mid-l+);
                     ULL h2 = cal(hb, l, mid, mid-l+);
                     if(h1 == h2)
                         l = mid;
                     else r = mid;
                 }
                 if(l != m-)
                     x++;
                 j = l+;
             }
             if(x <= k) {
                 ans = i;
                 break;
             }
         }
         printf("Case #%d: %d\n", t, ans);
     }
     return ;
 }