HDU 4455 Substrings(預處理+dp)

畫船聽雨發表於2014-11-29
原文網址 : https://blog.csdn.net/fulongxu/article/details/41592855

題目大意：給你n個數字，然後m次查詢，每次給你一個x，讓你求出來1到x，2到x+1。。。不同數的和。

需要各種預處理，處理出來所有的間隔之間有多少相同的數字，處理出來最後一個被去掉的間隔有多少個不重複的數字。

dp[i] = dp[i-1]-S+T.S代表最後被略去的那個區間的不同的數，T代表新區間擴張之後每個區間增加的不同的數的和。

Substrings

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1983    Accepted Submission(s): 608


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 a1,a2…an, 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<=106, 0<=Q<=104, 0<= a1,a2…an <=106
 

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



 





#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <time.h>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
///#define LL long long
#define LL __int64
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define mod 1000000007


using namespace std;

const int maxn = 1001000;

int num[maxn];
int vis[maxn];
int f[maxn];
int p[maxn];
LL dp[maxn];

int main()
{
    int n;
    while(~scanf("%d", &n) && n)
    {
        memset(vis, 0, sizeof(vis));
        memset(p, 0, sizeof(p));
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &num[i]);
            p[i-vis[num[i]]]++;
            vis[num[i]] = i;
        }
        memset(vis, 0, sizeof(vis));
        f[1] = 1;
        vis[num[n]] = 1;
        for(int i = 2; i <= n; i++)
        {
            if(vis[num[n-i+1]])
            {
                f[i] = f[i-1];
                continue;
            }
            vis[num[n-i+1]] = 1;
            f[i] = f[i-1]+1;
        }
        dp[1] = n;
        int sum = n;
        for(int i = 2; i <= n; i++)
        {
            dp[i] = dp[i-1]-f[i-1];
            sum -= p[i-1];
            dp[i] += sum;
        }
        int m;
        scanf("%d",&m);
        for(int i = 1; i <= m; i++)
        {
            int x;
            scanf("%d",&x);
            printf("%I64d\n",dp[x]);
        }
    }
    return 0;
}



            




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