hdu 5672 Strings 模拟

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
   
   
   
   

 

题目：给一个字符串，问有多少个子串至少含有k个不同的字母？

分析：从头开始扫一遍，如果遇见第一次出现过的，就计数cnt++，直到cnt==k，那么就从那个子串的开头开始删去字符，根据删除的字符出现的次数，做出相应的判断。

如果用map判断是否出现过会超时。

#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
using namespace std;
typedef long long ll;
const int N=1e6+9;
char s[N];
int k;
int vis[300];
int main()
{
    //freopen("f.txt","r",stdin);
    int T;scanf("%d",&T);
    while(T--){
        scanf("%s%d",s,&k);
        int slen=strlen(s);
        if(k==1){
            printf("%I64d\n",(ll)slen*(slen+1)/2);continue;
        }
        int head=0;
        ll ans=0;
        int cnt=0;
        memset(vis,0,sizeof(vis));
        for(int i=0;i<slen;i++){
            if(vis[s[i]]>0){
                vis[s[i]]++;
                continue;
            }
            vis[s[i]]=1;
            cnt++;
            if(cnt==k){
                while(head<i){
                    ans+=slen-i; 
                    if(vis[s[head]]==1){
                        vis[s[head]]=0;
                        head++;
                        cnt--;
                        break;
                    }
                    else{
                        vis[s[head]]--;head++;
                    }
                }
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}