HDU4455 Substrings（DP）

Posted 2022-06-02 松子茶

Substrings

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2918    Accepted Submission(s): 895
Problem Description
XXX has an array of length n. XXX wants to know that, for a given w, what is the sum of the distinct elements’ number in all substrings of length w. For example, the array is 1 1 2 3 4 4 5 When w = 3, there are five substrings of length 3. They are (1,1,2),(1,2,3),(2,3,4),(3,4,4),(4,4,5)
The distinct elements’ number of those five substrings are 2,3,3,2,2.
So the sum of the distinct elements’ number should be 2+3+3+2+2 = 12  

Input

There are several test cases.
Each test case starts with a positive integer n, the array length. The next line consists of n integers a ₁,a ₂…a _n, representing the elements of the array.
Then there is a line with an integer Q, the number of queries. At last Q lines follow, each contains one integer w, the substring length of query. The input data ends with n = 0 For all cases, 0<w<=n<=10 ⁶, 0<=Q<=10 ⁴, 0<= a ₁,a ₂…a _n <=10 ⁶

 Output

For each test case, your program should output exactly Q lines, the sum of the distinct number in all substrings of length w for each query.

 Sample Input

  
   7
1 1 2 3 4 4 5
3
1
2
3
0
  

 Sample Output

  
   7
10
12
  

 Source

2012 Asia Hangzhou Regional Contest   

这题不容易想到，一看题目，看到这数据范围，看到查询的方式。。。一直在往树状数组或者线段树方面去想。想到了用DP解决就不难了。用DP的思路O(n)复杂度解决。以样例为例说明：1 1 2 3 4 4 5；明显dp[1]=n=7;长度为1的时候有7个区间。从长度为1到长度为2，就是把前6个区间往后增加一个数，把最后一个区间去掉。增加的6个数要看在该区间是否出现过，只要看它上一个相等的元素距离是否大于2.所以dp[2]=dp[1]-1+4;   以此类推就可以得出所以的dp值了。dp[i]=dp[i-1]-A+B;减的A是最后一个长度为i-1的区间的不同数的个数，这个很容易预处理得出来。加的B是第t个数到它上一个数的距离大于i-1的个数.这个B值也容易得出。用s[i]表示离上一个数的距离为i的个数，不断减掉就得到B了。   具体看代码：

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
const int MAXN=1000010;
int a[MAXN];//1-n输入的数列
int f[MAXN];//f[i]表示a[i]在前面最近出现的位置,f[i]==0表示从左到右第一次出现
int s[MAXN];//s[i]表示 t-f[t]==i,1<=t<=n的t的个数,即离上一个相等元素的距离为i的个数
long long dp[MAXN];//需要输出的结果
int ss[MAXN];//ss[i]表示最后的i个数含有的不同元素的个数

int main()

    int n;
    int m;
    while(scanf("%d",&n)==1 && n)
    
        memset(f,0,sizeof(f));
        memset(s,0,sizeof(s));
        //顺着求s数组
        for(int i=1;i<=n;i++)
        
            scanf("%d",&a[i]);
            s[i-f[a[i]]]++;
            f[a[i]]=i;
        

        memset(f,0,sizeof(f));//f数组标记在后面是否出现过
        ss[1]=1;
        f[a[n]]=1;
        for(int i=2;i<=n;i++)
        
            if(f[a[n-i+1]]==0)
            
                f[a[n-i+1]]=1;
                ss[i]=ss[i-1]+1;
            
            else ss[i]=ss[i-1];
        
        dp[1]=n;
        int sum=n;
        //从dp[i-1]扩展到dp[i]就是去掉最后一个区间的个数，把前面的区间长度增加1，
        //加上相应增加的种类数
        for(int i=2;i<=n;i++)
        
            dp[i]=dp[i-1]-ss[i-1];//减掉最后一个区间的种类数
            sum-=s[i-1];
            dp[i]+=sum;//加上前面的区间增加一个长度后增加的种类数
        
        scanf("%d",&m);
        int t;
        while(m--)
        
            scanf("%d",&t);
            printf("%I64d\\n",dp[t]);
        
    
    return 0;