KMP POJ 3461 Oulipo-白红宇

KMP POJ 3461 Oulipo

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发布时间：2019-06-27

本文共 2958 字，大约阅读时间需要 9 分钟。

1 /* 2     题意：问一个串在另一个串出现的次数(可重复) 3     KMP：模板题 4 */ 5 /************************************************ 6 * Author        :Running_Time 7 * Created Time  :2015-8-9 19:45:40 8 * File Name     :POJ_3461.cpp 9  ************************************************/10 11 #include 
     
      12 #include 
      
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                   25 #include 
                   
                    26 #include 
                    
                     27 #include 
                     
                      28 using namespace std;29 30 #define lson l, mid, rt << 131 #define rson mid + 1, r, rt << 1 | 132 typedef long long ll;33 const int MAXN = 1e6 + 10;34 const int INF = 0x3f3f3f3f;35 const int MOD = 1e9 + 7;36 int nex[MAXN];37 char s[MAXN], t[MAXN];38 39 void get_nex(int lm) {40 int i = 0, j = -1; nex[0] = -1;41 while (i < lm) {42 if (j == -1 || t[j] == t[i]) {43 j++; i++; nex[i] = j;44 }45 else j = nex[j];46 }47 }48 49 int KMP(void) {50 int ln = strlen (s); int lm = strlen (t);51 get_nex (lm);52 int i = 0, j = 0; int ans = 0;53 while (i < ln) {54 while (j != -1 && t[j] != s[i]) j = nex[j];55 j++; i++;56 if (j >= lm) {57 ans++; j = nex[j];58 }59 }60 return ans; 61 }62 63 int main(void) { //POJ 3461 Oulipo64 int T; scanf ("%d", &T);65 while (T--) {66 scanf ("%s%s", t, s); 67 printf ("%d\n", KMP ());68 }69 70 return 0;71 }

1 /* 2     题目链接：http://oj.acm.zstu.edu.cn/JudgeOnline/problem.php?id=4194 3      给你两个字符串A，B，请输出B字符串在A字符串中出现了几次(不可重复) 4 */ 5 /************************************************ 6 * Author        :Running_Time 7 * Created Time  :2015-8-9 19:45:40 8 * File Name     :ZSTU_4194.cpp 9  ************************************************/10 11 #include 
      
       12 #include 
       
        13 #include 
        
         14 #include 
         
          15 #include 
          
           16 #include 
           
            17 #include 
            
             18 #include 
             
              19 #include 
              
               20 #include 
               
                21 #include 
                
                 22 #include 
                 
                  23 #include 
                  24 #include 
                   
                    25 #include 
                    
                     26 #include 
                     
                      27 #include 
                      
                       28 using namespace std;29 30 #define lson l, mid, rt << 131 #define rson mid + 1, r, rt << 1 | 132 typedef long long ll;33 const int MAXN = 1e6 + 10;34 const int INF = 0x3f3f3f3f;35 const int MOD = 1e9 + 7;36 int nex[MAXN];37 char s[MAXN], t[MAXN];38 39 void get_nex(int lm) {40 int i = 0, j = -1; nex[0] = -1;41 while (i < lm) {42 if (j == -1 || t[j] == t[i]) {43 i++; j++; nex[i] = j;44 }45 else j = nex[j];46 }47 }48 49 int KMP(void) {50 int ln = strlen (s);51 int lm = strlen (t);52 get_nex (lm);53 int i = 0, j = 0; int ans = 0;54 while (i < ln) {55 while (j != -1 && s[i] != t[j]) j = nex[j];56 i++; j++;57 if (j == lm) {58 ans++; j = 0; //改动这里就是重新匹配59 }60 }61 return ans; 62 }63 64 int main(void) {65 while (scanf ("%s%s", s, t) == 2) {66 printf ("%d\n", KMP ());67 }68 69 return 0;70 }71 72 不可重复的匹配