hdu 5672 String 尺取法

String

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


[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, output the number of substrings that contain at least k dictinct characters.

[align=left]Sample Input[/align]

2
abcabcabca
4
abcabcabcabc
3

[align=left]Sample Output[/align]

0
55

[align=left]Source[/align]
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;
}

View Code