传送门：Palindrome
题意：给定一个字符串，求最长回文子串。
分析：manach裸题，核心理解mx>i?p[i]=min(p[2*id-i],mx-i):1.
#pragma comment(linker,"/STACK:1024000000,1024000000") #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <limits.h> #include <iostream> #include <algorithm> #include <queue> #include <cstdlib> #include <stack> #include <vector> #include <set> #include <map> #define LL long long #define mod 1000000007 #define inf 0x3f3f3f3f #define eps 1e-6 #define N 1000010 #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define PII pair<int,int> using namespace std; inline LL read() { char ch=getchar();LL x=0,f=1; while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();} while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();} return x*f; } int p[N<<1],ans,len,num,mx,id; char s[N],str[N<<1]; void build() { len=strlen(s);num=0; str[num++]='@';str[num++]='#'; for(int i=0;i<len;i++) { str[num++]=s[i]; str[num++]='#'; } str[num]=0; } void manacher() { ans=0;mx=0; memset(p,0,sizeof(p)); for(int i=1;i<num;i++) { if(mx>i)p[i]=min(p[2*id-i],mx-i); else p[i]=1; while(str[i-p[i]]==str[i+p[i]])p[i]++; if(p[i]+i>mx)mx=p[i]+i,id=i; if(ans<p[i]-1)ans=p[i]-1; } } int main() { int cas=1; while(scanf("%s",s)>0) { if(strcmp(s,"END")==0)break; build(); manacher(); printf("Case %d: %d ",cas++,ans); } }