HDU5056 BoringCount--线性扫一遍

11754936

2014-09-29 10:08:45

Accepted

5056

31MS

392K

1257 B

G++

czy

好简单的思路，怎么就没想到呢。。。。。

Boring count

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 250    Accepted Submission(s): 98

Problem Description

You are given a string S consisting of lowercase letters, and your task is counting the number of substring that the number of each lowercase letter in the substring is no more than K.

 

Input

In the first line there is an integer T , indicates the number of test cases. For each case, the first line contains a string which only consist of lowercase letters. The second line contains an integer K.
[Technical Specification] 1<=T<= 100 1 <= the length of S <= 100000 1 <= K <= 100000

 

Output

For each case, output a line contains the answer.

 

Sample Input

3 abc 1 abcabc 1 abcabc 2

 

Sample Output

6 15 21

 

Source

BestCoder Round #11 (Div. 2)

 

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官方题解：

1003 Boring count
枚举字符串下标i，每次计算以i为结尾的符合条件的最长串。那么以i为结尾的符合条件子串个数就是最长串的长度。求和即可。
计算以i为结尾的符合条件的最长串两种方法：
1.维护一个起点下标startPos，初始为1。如果当前为i，那么cnt[str[i]]++，如果大于k的话，就while( str[startPos] != str[i+1] ) cnt[str[startPos]]--, startPos++; 每次都保证 startPos~i区间每个字母个数都不超过k个。ans += ( i-startPos+1 )。 时间复杂度O(n)
2.预处理出所有字母的前缀和。然后通过二分找出以i为结尾的符合条件的最长串的左边界。时间复杂度O(nlogn)，写的不够好的可能超时。

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<string>

#define N 100005
#define M 15
#define mod 10000007
//#define p 10000007
#define mod2 100000000
#define ll long long
#define LL long long
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b)

using namespace std;

int T;
int n;
ll l;
char s[N];
ll ans;
ll vis[30];
ll k;
ll te;

void ini()
{
    ans=0;
    scanf("%s",s);
    scanf("%I64d",&k);
    l=strlen(s);
    memset(vis,0,sizeof(vis));
}


void solve()
{
    ll i;
    ll pre;
    pre=0;
    for(i=0;i<l;i++){
        vis[ s[i]-'a' ]++;
        while(vis[ s[i]-'a' ]>k){
            vis[ s[pre]-'a' ]--;
            pre++;
        }
        ans+=i-pre+1;
    }
}

void out()
{
    printf("%I64d
",ans);
}

int main()
{
    //freopen("data.in","r",stdin);
    //freopen("data.out","w",stdout);
    scanf("%d",&T);
   // for(int ccnt=1;ccnt<=T;ccnt++)
    while(T--)
   // while(scanf("%d",&n)!=EOF)
    {
      //  if(n==0 && m==0) break;
        //printf("Case %d: ",ccnt);
        ini();
        solve();
        out();
    }

    return 0;
}