本文共 1128 字，大约阅读时间需要 3 分钟。
传送门：
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 #include #include #include #include #include #include #include #include #include #include #include #include #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 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 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 0) { if(strcmp(s,"END")==0)break; build(); manacher(); printf("Case %d: %d\n",cas++,ans); }}
转载于:https://www.cnblogs.com/lienus/p/4299443.html
