HDU2087,1686 KMP-白红宇

HDU2087,1686 KMP

阅读量：152 次

发布时间：2019-02-28

本文共 3378 字，大约阅读时间需要 11 分钟。

这两题都是统计一个字符串中另一个字符串的数量，只不过一题可以重叠，另一题不能重叠。

在代码上出了输入格式，也就是一个语句的区别，其实就是找到了一个文本串和模式串匹配的子串之后，是不是从模式串开头重新找的区别。

第一题AC代码：

/* * @Author: hesorchen * @Date: 2020-07-02 22:19:34 * @LastEditTime: 2020-07-11 12:49:56 * @Description: https://hesorchen.github.io/ */#include #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 using namespace std;#define endl '\n'#define PI acos(-1)#define PB push_back#define ll long long#define INF 0x3f3f3f3f#define mod 1000000007#define pll pair
                 
                  #define lowbit(abcd) (abcd & (-abcd))#define max(a, b) ((a > b) ? (a) : (b))#define min(a, b) ((a < b) ? (a) : (b))#define IOS \ ios::sync_with_stdio(false); \ cin.tie(0); \ cout.tie(0);#define FRE \ { \ freopen("in.txt", "r", stdin); \ freopen("out.txt", "w", stdout); \ }inline ll read(){ ll x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar(); } return x * f;}//==============================================================================ll nxt[1000010];char a[1000010], b[1000010];void get_nxt(){ ll lenb = strlen(b); int j = 0, k = -1; nxt[0] = -1; while (j < lenb) { if (k == -1 || b[j] == b[k]) nxt[++j] = ++k; else k = nxt[k]; }}int KMP(){ ll lenb = strlen(b), lena = strlen(a), res = 0; get_nxt(); ll i = 0, j = 0; while (i < lena) { if (j == -1 || a[i] == b[j]) i++, j++; else j = nxt[j]; if (j == lenb) { res++; // 能不能重复 // j = nxt[j]; j = 0; } } return res;}int main(){ IOS; ll n; while (cin >> a) { if (a[0] == '#') break; cin >> b; cout << KMP() << endl; } return 0;}/*ababaaba*/

第二题AC代码：

/* * @Author: hesorchen * @Date: 2020-07-02 22:19:34 * @LastEditTime: 2020-07-11 12:58:25 * @Description: https://hesorchen.github.io/ */#include #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 using namespace std;#define endl '\n'#define PI acos(-1)#define PB push_back#define ll long long#define INF 0x3f3f3f3f#define mod 1000000007#define pll pair
                 
                  #define lowbit(abcd) (abcd & (-abcd))#define max(a, b) ((a > b) ? (a) : (b))#define min(a, b) ((a < b) ? (a) : (b))#define IOS \ ios::sync_with_stdio(false); \ cin.tie(0); \ cout.tie(0);#define FRE \ { \ freopen("in.txt", "r", stdin); \ freopen("out.txt", "w", stdout); \ }inline ll read(){ ll x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar(); } return x * f;}//==============================================================================ll nxt[1000010];char a[1000010], b[1000010];void get_nxt(){ ll lenb = strlen(b); int j = 0, k = -1; nxt[0] = -1; while (j < lenb) { if (k == -1 || b[j] == b[k]) nxt[++j] = ++k; else k = nxt[k]; }}int KMP(){ ll lenb = strlen(b), lena = strlen(a), res = 0; get_nxt(); ll i = 0, j = 0; while (i < lena) { if (j == -1 || a[i] == b[j]) i++, j++; else j = nxt[j]; if (j == lenb) { res++; // 能不能重复 j = nxt[j]; // j = 0; } } return res;}int main(){ IOS; ll n; cin >> n; while (n--) { if (a[0] == '#') break; cin >> b >> a; cout << KMP() << endl; } return 0;}/*ababaaba*/

转载地址：http://igod.baihongyu.com/

你可能感兴趣的文章