hdu 5672 String 尺取法

String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

There is a string S.S only contain lower case English character.(10≤length(S)≤1,000,000)
How many substrings there are that contain at least k(1≤k≤26) distinct characters?

 

 

Input

There are multiple test cases. The first line of input contains an integer T(1≤T≤10) indicating the number of test cases. For each test case:

The first line contains string S.
The second line contains a integer k(1≤k≤26).

 

 

Output

For each test case, output the number of substrings that contain at least k dictinct characters.

 

 

Sample Input

2 abcabcabca 4 abcabcabcabc 3

 

 

Sample Output

0 55

 

 

Source

BestCoder Round #81 (div.2)

思路：就是从左端点找到最近的那个符合条件的右端点；这题特容易超时

尺取法：http://wenku.baidu.com/link?url=_jFXiTHG4ZN60Ki0U5Svb26oKLbbUMtJAwrSnkDC1W1e9RqFK_DaolSUE3MyCKmrv2oGEKWn_GN5P7IQuV0Qp5jxA1SApZHSBYI4NqEYq_u

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#define true ture
#define false flase
using namespace std;
#define ll __int64
#define inf 0xfffffff
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF )  return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
int flag[30];
char a[1000010];
int main()
{
    int x,y,z,i,t;
    scanf("%d",&x);
    while(x--)
    {
        memset(flag,0,sizeof(flag));
        scanf("%s",a);
        scanf("%d",&y);
        int st=0,en=0,ji=0;
        ll ans=0;
        int len=strlen(a);
        while(1)
        {
            while(en<len&&ji<y)
            {
                if(flag[a[en]-'a']==0)
                ji++;
                flag[a[en]-'a']++;
                en++;
            }
            if(ji<y)break;
            ans+=(len-en+1);
            flag[a[st]-'a']--;
            if(flag[a[st]-'a']==0)
            ji--;
            st++;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}