BZOJ2213: [Poi2011]Difference

2213: [Poi2011]Difference

Time Limit: 10 Sec  Memory Limit: 32 MB
Submit: 343  Solved: 108
[Submit][Status]

Description

A word consisting of lower-case letters of the English alphabet ('a'-'z') is given. We would like to choose a non-empty contiguous (i.e. one-piece) fragment of the word so as to maximise the difference in the number of occurrences of the most and the least frequent letter in the fragment. We are assuming that the least frequent letter has to occur at least once in the resulting fragment. In particular, should the fragment contain occurrences of only one letter, then the most and the least frequent letter in it coincide.

已知一个长度为n的由小写字母组成的字符串，求其中连续的一段，满足该段中出现最多的字母出现的个数减去该段中出现最少的字母出现的个数最大。求这个个数。

Input

The first line of the standard input holds one integer (1<=N<=1000000)() that denotes the length of the word. The second line holds a word consisting of lower-case letters of the English alphabet.

第一行，n
第二行，该字符串
1<=n<=1000000

Output

The first and only line of the standard output is to hold a single integer, equal to the maximum difference in the number of occurrences of the most and the least frequent letter that is attained in some non-empty contiguous fragment of the input word.

一行，表示结果

Sample Input

10
aabbaaabab

Sample Output

3
Explanation of the example: The fragment that attains the difference of 3 in the number of occurrences of a and b is aaaba.

HINT

 

Source

题解：

copy：

记f[i][j]表示第i位j字母出现的次数，则ans=max(f[i1][a]-f[i0][a]-(f[i1][b]-f[i0][b]))=max(f[i1][a]-f[i1][b]-(f[i0][a]-f[i0][b]))

所以维护一下f[i][a]-f[i][b]的最小值，更新的时候看一下。

连续一段区间只有一个字母：记录次小值，用上一次更新答案时的cnt[a]或者cnt[b]来判断即可。

其实这题和连续最大和差不多，我太sb了想不到。。。

代码理解起来好困难，看了好久stupid_lulu's 的代码没看懂，唉，挖个坑，以后来填

代码：

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<string>
 #define inf 1000000000
 #define maxn 500+100
 #define maxm 500+100
 #define eps 1e-10
 #define ll long long
 #define pa pair<int,int>
 #define for0(i,n) for(int i=0;i<=(n);i++)
 #define for1(i,n) for(int i=1;i<=(n);i++)
 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 #define mod 1000000007
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}
     return x*f;
 }
 int n,ans=,f[][],g[][];
 int main()
 {
     freopen("input.txt","r",stdin);
     freopen("output.txt","w",stdout);
     n=read();
     for1(i,n)
      {
          char ch=getchar();
          while(ch>'z'||ch<'a')ch=getchar();
          int x=ch-'a';
          for0(j,)
           {
              f[x][j]++;
              f[j][x]--;
              if(!g[j][x])g[j][x]=;
              if(g[j][x]==){g[j][x]=;f[j][x]++;}
              if(f[j][x]<=-){f[j][x]=-;g[j][x]=;}
              if(g[j][x])ans=max(ans,f[j][x]);
              if(g[x][j])ans=max(ans,f[x][j]);
           }
      }
     printf("%d\n",ans);
     return ;
 }