博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
Ural 2040. Palindromes and Super Abilities 2 回文自动机
阅读量:5775 次
发布时间:2019-06-18

本文共 8889 字,大约阅读时间需要 29 分钟。

2040. Palindromes and Super Abilities 2

题目连接:

Description

Dima adds letters s1, …, sn one by one to the end of a word. After each letter, he asks Misha to tell him how many new palindrome substrings appeared when he added that letter. Two substrings are considered distinct if they are different as strings. Which n numbers will be said by Misha if it is known that he is never wrong?

Input

The input contains a string s1 … sn consisting of letters ‘a’ and ‘b’ (1 ≤ n ≤ 5 000 000).

Output

Print n numbers without spaces: i-th number must be the number of palindrome substrings of the prefix s1 … si minus the number of palindrome substrings of the prefix s1 … si−1. The first number in the output should be one.

Sample Input

abbbba

Sample Output

111111

Hint

题意

每次往后增加一个字符的时候,如果本质不同的回文串增加了,那就输出1,否则输出0

题解:

回文自动机/回文树裸题,卡了输入输出

代码

#include 
using namespace std;const int maxn = 5e6 + 15;namespace fastIO{ #define BUF_SIZE 100000 #define OUT_SIZE 100000 #define ll long long //fread->read bool IOerror=0; inline char nc(){ static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE; if (p1==pend){ p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin); if (pend==p1){IOerror=1;return -1;} //{printf("IO error!\n");system("pause");for (;;);exit(0);} } return *p1++; } inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';} inline void read(int &x){ bool sign=0; char ch=nc(); x=0; for (;blank(ch);ch=nc()); if (IOerror)return; if (ch=='-')sign=1,ch=nc(); for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; if (sign)x=-x; } inline void read(ll &x){ bool sign=0; char ch=nc(); x=0; for (;blank(ch);ch=nc()); if (IOerror)return; if (ch=='-')sign=1,ch=nc(); for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; if (sign)x=-x; } inline void read(double &x){ bool sign=0; char ch=nc(); x=0; for (;blank(ch);ch=nc()); if (IOerror)return; if (ch=='-')sign=1,ch=nc(); for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; if (ch=='.'){ double tmp=1; ch=nc(); for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0'); } if (sign)x=-x; } inline void read(char *s){ char ch=nc(); for (;blank(ch);ch=nc()); if (IOerror)return; for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch; *s=0; } inline void read(char &c){ for (c=nc();blank(c);c=nc()); if (IOerror){c=-1;return;} } //getchar->read inline void read1(int &x){ char ch;int bo=0;x=0; for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1; for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar()); if (bo)x=-x; } inline void read1(ll &x){ char ch;int bo=0;x=0; for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1; for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar()); if (bo)x=-x; } inline void read1(double &x){ char ch;int bo=0;x=0; for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1; for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar()); if (ch=='.'){ double tmp=1; for (ch=getchar();ch>='0'&&ch<='9';tmp/=10.0,x+=tmp*(ch-'0'),ch=getchar()); } if (bo)x=-x; } inline void read1(char *s){ char ch=getchar(); for (;blank(ch);ch=getchar()); for (;!blank(ch);ch=getchar())*s++=ch; *s=0; } inline void read1(char &c){for (c=getchar();blank(c);c=getchar());} //scanf->read inline void read2(int &x){scanf("%d",&x);} inline void read2(ll &x){ #ifdef _WIN32 scanf("%I64d",&x); #else #ifdef __linux scanf("%lld",&x); #else puts("error:can't recognize the system!"); #endif #endif } inline void read2(double &x){scanf("%lf",&x);} inline void read2(char *s){scanf("%s",s);} inline void read2(char &c){scanf(" %c",&c);} inline void readln2(char *s){gets(s);} //fwrite->write struct Ostream_fwrite{ char *buf,*p1,*pend; Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;} void out(char ch){ if (p1==pend){ fwrite(buf,1,BUF_SIZE,stdout);p1=buf; } *p1++=ch; } void print(int x){ static char s[15],*s1;s1=s; if (!x)*s1++='0';if (x<0)out('-'),x=-x; while(x)*s1++=x%10+'0',x/=10; while(s1--!=s)out(*s1); } void println(int x){ static char s[15],*s1;s1=s; if (!x)*s1++='0';if (x<0)out('-'),x=-x; while(x)*s1++=x%10+'0',x/=10; while(s1--!=s)out(*s1); out('\n'); } void print(ll x){ static char s[25],*s1;s1=s; if (!x)*s1++='0';if (x<0)out('-'),x=-x; while(x)*s1++=x%10+'0',x/=10; while(s1--!=s)out(*s1); } void println(ll x){ static char s[25],*s1;s1=s; if (!x)*s1++='0';if (x<0)out('-'),x=-x; while(x)*s1++=x%10+'0',x/=10; while(s1--!=s)out(*s1); out('\n'); } void print(double x,int y){ static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000, 1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL, 100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL}; if (x<-1e-12)out('-'),x=-x;x*=mul[y]; ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1; ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2); if (y>0){out('.'); for (size_t i=1;i
write char Out[OUT_SIZE],*o=Out; inline void print1(int x){ static char buf[15]; char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x; while(x)*p1++=x%10+'0',x/=10; while(p1--!=buf)*o++=*p1; } inline void println1(int x){print1(x);*o++='\n';} inline void print1(ll x){ static char buf[25]; char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x; while(x)*p1++=x%10+'0',x/=10; while(p1--!=buf)*o++=*p1; } inline void println1(ll x){print1(x);*o++='\n';} inline void print1(char c){*o++=c;} inline void println1(char c){*o++=c;*o++='\n';} inline void print1(char *s){while (*s)*o++=*s++;} inline void println1(char *s){print1(s);*o++='\n';} inline void println1(){*o++='\n';} inline void flush1(){if (o!=Out){if (*(o-1)=='\n')*--o=0;puts(Out);}} struct puts_write{ ~puts_write(){flush1();} }_puts; inline void print2(int x){printf("%d",x);} inline void println2(int x){printf("%d\n",x);} inline void print2(char x){printf("%c",x);} inline void println2(char x){printf("%c\n",x);} inline void print2(ll x){ #ifdef _WIN32 printf("%I64d",x); #else #ifdef __linux printf("%lld",x); #else puts("error:can't recognize the system!"); #endif #endif } inline void println2(ll x){print2(x);printf("\n");} inline void println2(){printf("\n");} #undef ll #undef OUT_SIZE #undef BUF_SIZE};using namespace fastIO;char str[maxn] , output[maxn];struct Palindromic_Auto{ const static int LetterSize = 2; // 字符集大小 const static int TrieSize = maxn; // 可能的所有节点总数量 int tot; // 节点总数 int suffixlink[TrieSize]; // 后缀链接 struct node{ int ptr[LetterSize]; // 节点指针 int len ; // 长度 }tree[TrieSize]; inline int GetLetterIdx(char c){ // 获取字符哈希,在[0,LetterSize)内 return c - 'a'; } inline void init_node(node & x , int len = 0){ memset( x.ptr , 0 , sizeof( x.ptr ) ); x.len = len ; } void insert( const char * str ){ int len = strlen( str ); int j = 0 ; for(int i = 0 ; i < len ; ++ i){ int idx = GetLetterIdx( str[i] ); for( ; i-tree[j].len-1 < 0 || str[i]!=str[i-tree[j].len-1] ; j=suffixlink[j]); //Max Math for suffixlink if(tree[j].ptr[idx]){ j = tree[j].ptr[idx]; output[i] = '0'; continue; } output[i] = '1'; int cur = j; init_node( tree[tot] , tree[cur].len + 2 ); tree[cur].ptr[idx]=tot,j=tot++; if( tree[j].len == 1 ){ suffixlink[j] = 1; continue; } for( cur = suffixlink[cur] ; i - tree[cur].len - 1 < 0 || str[i]!= str[i-tree[cur].len-1] ; cur = suffixlink[cur] ); suffixlink[j]=tree[cur].ptr[idx]; } } void init(){ tot = 0 ; init_node( tree[tot ++ ] , -1 ); init_node( tree[tot ++ ] , 0 ); }}pa_auto;int main(int argc,char *argv[]){ read(str); pa_auto.init(); pa_auto.insert( str ); println(output); return 0;}

转载地址:http://crhux.baihongyu.com/

你可能感兴趣的文章
详解数据结构中的“数组”与编程语言中的“数组”的区别和联系
查看>>
android判断状态栏是否可见
查看>>
微软一年封锁50亿封钓鱼邮件 提供企业1.4亿次安全建议
查看>>
Java集合系列(一)Iterable & Iterator
查看>>
SSM搭建博客系统(三):数据库设计
查看>>
mybatis使用
查看>>
【坑】spring-boot-maven-plugin的executable配置
查看>>
原生js实现点名册效果
查看>>
(一)C#关于变量和类型的冷知识
查看>>
ios 基础问题记录
查看>>
层叠上下文与Z-index学习笔记
查看>>
【Java并发】synchronized
查看>>
八大排序算法实战:思想与实现
查看>>
【OSX】解决Terminal ssh连接"Write failed Broken pipe"问题
查看>>
聊聊散列表以及HashMap内部实现
查看>>
JS函数柯里化
查看>>
Spring AOP注解通过@Autowired,@Resource,@Qualifier,@PostConstruct,@PreDestroy注入属性的...
查看>>
html之iframe,a
查看>>
flutter 混合开发 (android 实际操作)
查看>>
盒模型的一些介绍
查看>>