2519: Find the longest section 位运算+前缀和

寻找最长偶数串段

最新推荐文章于 2023-11-14 14:48:15 发布

原创最新推荐文章于 2023-11-14 14:48:15 发布 · 685 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#output #string #each #struct #input #up

joj acm_数学问题专栏收录该内容

63 篇文章

订阅专栏

本文介绍了一种算法，用于解决寻找多个字符串中最长连续子串的问题，其中子串的所有字符出现次数均为偶数。该算法通过位运算来快速判断字符串中字符的奇偶性，并利用排序技巧找到符合条件的最长子串。

2519: Find the longest section

There are n strings with the length of no more than 100 characters. Can you find the longest continuous string section(ie. S(i)S(i+1)...S(j) and j-i+1 is maximaze) that all the letters appear even times.

Input

There will be multiple test case. For each test: The first line has a positive number n(1 <= n <= 10000). Then followed n rows, each row has a string making up with lowercase letters.

Output

Output the longest section(left, right),where left is the beginning loction of the section and right is the ending loction of the section. If there is more than one such section, output the one whose left is smallest.If there is none such section,just output "No one".

Sample Input

3
a
ab
b
7
a
ab
b
e
wang
wang
wangwang
2
mars
wang

Sample Output

1 3
1 3
No one

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int maxn=10000+100;
char str[300];
struct Node
{
int sum;
int index;
};
Node g[maxn];
bool cmp(Node h,Node k)
{
if(h.sum!=k.sum) return h.sum<k.sum;
return h.index<k.index;
}
int main()
{
int n;
while(scanf("%d",&n)==1)
{
  for(int i=0;i<=n;i++) g[i].sum=0,g[i].index=i;
  for(int i=1;i<=n;i++)
  {
   scanf("%s",str);
   int cnt=0;
   for(int j=0;str[j];j++)
   {
    cnt=cnt^(1<<(str[j]-'a'));
   }
   g[i].sum=g[i-1].sum^cnt;
  }
  sort(g,g+n+1,cmp);
  int flag=0,st=-1,ed=-1,_max=0;
  for(int i=0;i<n;)
  {
   if(g[i].sum==g[i+1].sum)
   {
    flag=1;
    int j=i+1;
    while(j<=n&&g[i].sum==g[j].sum) j++;
    int len=g[j-1].index-g[i].index;
    if(len>_max) _max=len,st=g[i].index,ed=g[j-1].index;
    else if(len==_max)
    {
     if(st>g[i].index) st=g[i].index,ed=g[j-1].index;
    }
    i=j;
   }
   else
   {
    i++;
   }
  }
  if(!flag) printf("No one/n");
  else printf("%d %d/n",st+1,ed);
}
return 0;
}