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给定一个字符串,把其中出现的 A串替换为 B串
KMP入门题,对原串匹配A串,跑一遍KMP
然后匹配到终点的时候替换就好了
最后再输出替换的结果
时间复杂度 O(N)
#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;
typedef pair<int,int> Pii;
typedef long long LL;
typedef unsigned long long ULL;
typedef double DBL;
typedef long double LDBL;
#define MST(a,b) memset(a,b,sizeof(a))
#define CLR(a) MST(a,0)
#define Sqr(a) (a*a)
const int maxn=5e6+10;
struct KMP
{
char *patn;
int plen,*fail;
KMP(char*,int);
~KMP(){delete[] fail;}
int match(char*,int,char*,int,char*);
};
char inpt[maxn];
char patn[maxn];
char conv[maxn];
char ans[maxn];
int main()
{
int T;
scanf("%d", &T);
for(int ck=1; ck<=T; ck++)
{
scanf(" %s %s %s", inpt, patn, conv);
KMP kmp(patn,strlen(patn));
CLR(ans);
kmp.match(inpt,strlen(inpt),conv,strlen(conv),ans);
puts(ans);
}
return 0;
}
KMP::KMP(char tpatn[],int len):patn(tpatn),plen(len)
{
fail=new int[plen+1];
fail[0]=fail[1]=0;
for(int i=1; i<plen; i++)
{
int fp = fail[i];
while(fp && patn[fp] != patn[i]) fp=fail[fp];
if(patn[fp]==patn[i]) fail[i+1] = fp+1;
else fail[i+1] = 0;
}
}
int KMP::match(char str[], int len, char conv[], int clen, char res[])
{
int np=0,end=0;
for(int i=0; i<len; i++)
{
res[end++]=inpt[i];
while( np && patn[np] != str[i]) np = fail[np];
if( patn[np] == str[i] ) np++;
if( np == plen )
{
np=0;
for(int j=end-plen; j<end-plen+clen; j++) res[j]=conv[j-(end-plen)];
end=end-plen+clen;
}
}
res[end]=0;
return end;
}
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原文地址:http://blog.csdn.net/u012015746/article/details/51366852