BZOJ2084:[POI2010]Antisymmetry

浅谈\(Manacher\)https://www.cnblogs.com/AKMer/p/10431603.html

题目传送门:https://lydsy.com/JudgeOnline/problem.php?id=2084

题目求的就是偶数长度回文串个数。不过匹配从相等变成了异或等于\(1\),在\(Manacher\)算法上稍作改进即可。

时间复杂度:\(O(n)\)

空间复杂度:\(O(n)\)

代码如下:

#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn=1e6+5;

int n,ans;
int p[maxn];
char s[maxn];

int read() {
	int x=0,f=1;char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
	return x*f;
}

int main() {
	n=read(),scanf("%s",s+1);
	for(int i=n;i;i--)
		s[i<<1]=s[i],s[(i<<1)-1]='#';
	s[0]='$',s[n<<1|1]='#',n=n<<1|1;
	int id=0,mx=0;
	for(int i=1;i<=n;i++) {
		p[i]=i<=mx?min(mx-i+1,p[(id<<1)-i]):1;
		while(s[i-p[i]]-'0'+s[i+p[i]]-'0'==1||s[i-p[i]]=='#')p[i]++;
		if(s[i]=='#')ans+=(p[i]-1)/2;
	}
	printf("%d\n",ans);
	return 0;
}
posted @ 2019-02-26 15:35  AKMer  阅读(98)  评论(0编辑  收藏  举报