• hihoCoder 1015 KMP算法


    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
    ", 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
    ",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
    ",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
    ",m);
    84         next(m);
    85         n=strlen(s2+1);
    86         printf("%d
    ",kmp(n));
    87     }
    88     return 0;
    89 }
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  • 原文地址:https://www.cnblogs.com/nowandforever/p/4580395.html
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