#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <cstring>
#include <list>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
#define PF(x) (scanf("%d",&x))
#define PT(x,y) (scanf("%d%d",&x,&y))
#define PR(x) (printf("%d\n",x))
#define PRT(x,y)(printf("%d %d\n",x,y))
#define PB(x)(scanf("%I64d",&x))
#define PRB(x)(printf("%I64d\n",(x)))
typedef __int64 LL;
#define N 2000025
#define M 105
#define Mod 1000
#define Inf 0x7fffffff
char s[N];
char st[N];
int rad[N];
int len;
inline int maxx(int a,int b) {return a>b?a:b;}
inline int minx(int a,int b) {return a>b?b:a;}
void convert()
{
int i;
st[0] = '@';
st[1] = '#';
len = strlen(s);
for(i=0;i<len;i++)
{
st[2*i+2] = s[i];
st[2*i+3] = '#';
}
st[2*i+2]='\0';
}
int cal()
{
convert();
int j,k;
int mx = 0,id;
int ans = 0;
for(int i = 0;i<2*len+2;i++)// 使用strlen(st)超时,想想应该的
{
if(mx>i) rad[i] = minx(rad[2*id-i],mx-i);
else rad[i] = 1;
for(;st[i-rad[i]] == st[i+rad[i]];rad[i]++)
if(rad[i]+i>mx)
{
mx = rad[i] +i;
id = i;
}
if(ans<rad[i]) ans = rad[i];
}
return ans-1;
}
void init()
{
int test= 1;
while(scanf("%s",s)==1)
{
if(strcmp(s,"END")==0) break;
printf("Case %d: ",test);test++;
PR(cal());
}
return ;
}
int main()
{
init();
return 0;
}
poj 3974 最长回文串O(N)的算法

最新推荐文章于 2019-07-30 19:48:16 发布
