最小表示+kmp

#include<iostream>
using namespace std;
const int maxn=2000005;
//最小表示+最大表示+KMP
//問題描述，求字典序最小的和字典序最大的位置，若多個取最左邊的，并給出在這n個串中出現次數
//注意自身得情況，由于出始擴大了一倍，但自身只能算一次，這里要注意
char b[maxn];
char a[maxn];
int next[maxn];
int minshow(char *s)
{
??? int i=0,j=1,k=0,n=strlen(s);
??? while(i<n/2&&j<n/2&&k<n/2)
??? {
????????? if(s[i+k]==s[j+k])k++;
????????? else
????????? {
?????????????? if(s[i+k]>s[j+k])i+=(k+1);
?????????????? else j+=(k+1);
?????????????? if(i==j)j++;
?????????????? k=0;
?????????????? }???????????????????
????????? }
??? return min(i,j);
??? }
int maxshow(char *s)
{
??? int i=0,j=1,k=0,n=strlen(s);
??? while(i<n/2&&j<n/2&&k<n/2)
??? {
????????? if(s[i+k]==s[j+k])k++;
????????? else
????????? {
?????????????? if(s[i+k]<s[j+k])i+=(k+1);
?????????????? else j+=(k+1);
?????????????? if(i==j)j++;
?????????????? k=0;
?????????????? }???????????????????
????????? }
??? return min(i,j);
??? }
void slove(char *s)
{
??? int i,j,len=strlen(s);
??? next[0]=0;j=0;
??? for(i=1;i<len;i++)
??? {
??????? while(j>0&&s[j]!=s[i])j=next[j-1];
??????? if(s[j]==s[i])j++;
??????? next[i]=j;?????????????
??????? }
??? }???
int kmp(char *a,char *b)
{
??? int i,j=0,lena=strlen(a),lenb=strlen(b),c=0;
??? for(i=0;i<lenb;i++)
??? {
??????? while(j>0&&a[j]!=b[i])j=next[j-1];
??????? if(a[j]==b[i])j++;
??????? if(j==lena){j=next[j-1];c++;}??????????????
??????? }
??? return c;???
??? }???
int main()
{
??? int i,j,k,n,mins,maxs,mint,maxt;
??? while(gets(b))
??? {
????????? n=strlen(b);
????????? for(i=0;i<n;i++)b[i+n]=b[i];
????????? b[n+n]=0;
????????? mins=minshow(b)+1;
????????? for(i=0;i<n;i++)a[i]=b[i+mins-1];
????????? a[n]=0;
????????? slove(a);
????????? mint=kmp(a,b+1);
?????????
????????? maxs=maxshow(b)+1;
????????? for(i=0;i<n;i++)a[i]=b[i+maxs-1];
????????? a[n]=0;
????????? slove(a);
????????? maxt=kmp(a,b+1);
????????? printf("%d %d %d %d\n",mins,mint,maxs,maxt);???????
????????? }
??? return 0;
??? }

轉載于:https://www.cnblogs.com/zhaozhe/archive/2011/08/10/2133351.html

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