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http://hihocoder.com/problemset/problem/1015
因为kmp算法只预处理b串,因而适合求解,给定一个串和一群不同的a串,问b是那些a串的子串.
或者查找模式串在原串出现了几次.
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <vector> 5 #include <cstring> 6 #include <string> 7 #include <algorithm> 8 #include <string> 9 #include <set> 10 #include <functional> 11 #include <numeric> 12 #include <sstream> 13 #include <stack> 14 #include <map> 15 #include <queue> 16 #pragma comment(linker, "/STACK:102400000,102400000") 17 #define CL(arr, val) memset(arr, val, sizeof(arr)) 18 19 #define ll long long 20 #define inf 0x7f7f7f7f 21 #define lc l,m,rt<<1 22 #define rc m + 1,r,rt<<1|1 23 #define pi acos(-1.0) 24 25 #define L(x) (x) << 1 26 #define R(x) (x) << 1 | 1 27 #define MID(l, r) (l + r) >> 1 28 #define Min(x, y) (x) < (y) ? (x) : (y) 29 #define Max(x, y) (x) < (y) ? (y) : (x) 30 #define E(x) (1 << (x)) 31 #define iabs(x) (x) < 0 ? -(x) : (x) 32 #define OUT(x) printf("%I64d\n", x) 33 #define lowbit(x) (x)&(-x) 34 #define Read() freopen("a.txt", "r", stdin) 35 #define Write() freopen("b.txt", "w", stdout); 36 #define maxn 1000000000 37 #define N 2510 38 #define mod 1000000000 39 using namespace std; 40 41 int p[10010]; 42 char s1[10010],s2[1000010]; 43 int n,m; 44 void next(int m) 45 { 46 int j=0; 47 p[1]=0; 48 for(int i=2;i<=m;i++) 49 { 50 while(j>0&&(s1[j+1]!=s1[i])) j=p[j]; 51 if(s1[j+1]==s1[i]) j+=1; 52 p[i]=j; 53 //printf("%d\n",p[i]); 54 } 55 } 56 57 int kmp(int n) 58 { 59 int j=0,k=0; 60 for(int i=1;i<=n;i++) 61 { 62 while(j>0&&s1[j+1]!=s2[i]) j=p[j]; 63 if(s1[j+1]==s2[i]) j=j+1; 64 if(j==m) 65 { 66 k++; 67 j=p[j]; 68 } 69 //printf("%d\n",j); 70 } 71 return k; 72 } 73 int main() 74 { 75 // freopen("a.txt","r",stdin); 76 int k; 77 scanf("%d",&k); 78 for(int i=0;i<k;i++) 79 { 80 scanf("%s",s1+1); 81 scanf("%s",s2+1); 82 m=strlen(s1+1); 83 //printf("%d\n",m); 84 next(m); 85 n=strlen(s2+1); 86 printf("%d\n",kmp(n)); 87 } 88 return 0; 89 }
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原文地址:http://www.cnblogs.com/nowandforever/p/4580395.html