Ural1297palindrome (suffix array +lcp)

Find the longest palindrome substring, and this problem requires the first appearance, SA+LCP solve. Original idea: First inversion, splicing the original string and the reverse string, and then the longest common prefix of the two, the only thing to note is if there are more than one group to output the first group to appear. The original idea was wrong. Counter example: Zzzdzaadzzz. Can find their own, really not, to the Great God blog under the reading is also possible. Transmission Door

Positive solution: The first thing to do is to insert a special, impossible character between itself and the mirror. We enumerate the characters in each of the original strings and find them centered (in odd and even two cases), for example, for the characters with subscript I, when the palindrome string is odd, we are asking for the longest common prefix of the suffix of I and the 2*n-i suffix, Then according to the property of the height array into the minimum value of height[i+1]...height[2*n-i], then we think of using RMQ to find the most value of the interval.

#include <cstdio> #include <cstring> #define N (char s[n],str[n];
int rank[n<<1],rank1[n],sa[n],count[n],tmp[n],h[n],st[n][14],log[n]; inline int swap (int &x,int &y) {int t=x;x=y;y=t;} inline int min (int x,int y) {return x<y?x:y;} inline int rmq (in
	T l,int r) {int t=log[r-l+1];
return min (st[l][t],st[r-(1<<t) +1][t]);
	} inline int LCP (int x,int y) {int l=rank[x],r=rank[y];
	if (l>r) swap (l,r); l++;
Return RMQ (L,R);
	} int main () {//Freopen ("A.in", "R", stdin);
	while (scanf ("%s", str+1)!=eof) {int Ans=0,ansl=0,n=0,len=strlen (str+1);
	for (int i=1;i<=len;++i) s[++n]=str[i];
	S[++n]= ' z ' +1;for (int i=len;i>=1;--i) s[++n]=str[i];
	memset (rank,0,sizeof (rank));
	memset (Count,0,sizeof (count));
	memset (h,0,sizeof (h));
	for (int i=1;i<=n;++i) count[s[i]]=1;
	for (int i=1;i<=200;++i) count[i]+=count[i-1];
	for (int i=n;i>=1;--i) rank[i]=count[s[i]];
	int k=0;
		for (int p=1;k!=n;p<<=1) {memset (count,0,sizeof (count)); for (int i=1;i<=n;++i) count[rank[i+p]]++;
		for (int i=1;i<=n;++i) count[i]+=count[i-1];
		for (int i=n;i>=1;--i) tmp[count[rank[i+p]]--]=i;
		memset (Count,0,sizeof (count));
		for (int i=1;i<=n;++i) count[rank[tmp[i]]]++;
		for (int i=1;i<=n;++i) count[i]+=count[i-1];
		for (int i=n;i>=1;--i) sa[count[rank[tmp[i]]]--]=tmp[i];
		memcpy (rank1,rank,sizeof (RANK1));
		rank[sa[1]]=k=1; for (int i=2;i<=n;++i) {if (rank1[sa[i]]!=rank1[sa[i-1]]| |
			RANK1[SA[I]+P]!=RANK1[SA[I-1]+P]) ++k;
		Rank[sa[i]]=k;
	}}k=0;
		for (int i=1;i<=n;++i) {if (rank[i]==1) {h[1]=0;continue;} if (i==1| |
		H[rank[i-1]]<=1) k=0;
		if (k)--k;
		while (S[i+k]==s[sa[rank[i]-1]+k]) ++k;
	H[rank[i]]=k;
	}log[0]=-1;
	for (int i=1;i<=n;++i) log[i]=log[i>>1]+1;
	for (int i=1;i<=n;++i) st[i][0]=h[i]; for (int i=1;i<=log[n];++i) for (int j=1;j<=n;++j) if (j+ (1<<i-1) <=n) st[j][i]=min (st[j][i-1],st[j+ (1
			<<i-1)][i-1]);
	else break; for (int i=1;i<=len;++i) {//s[i] The longest palindrome/palindrome length is odd intX=LCP (N-i+1,i), a=i-x+1;x= (x<<1)-1;
		if (X>ans) ans=x,ansl=a;
		else if (X==ans) ansl=min (ansl,a);
		Palindrome length is even if (i==1) continue;
		X=LCP (n-i+2,i);a=i-x;x<<=1;
		if (X>ans) ans=x,ansl=a;
	else if (X==ans) ansl=min (ansl,a);
	} for (int i=0;i<ans;++i) Putchar (S[ansl+i]);
	Puts ("");}
return 0; }