light oj 1122 - Digit Count（dp）

最新推荐文章于 2018-08-14 16:45:04 发布

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最新推荐文章于 2018-08-14 16:45:04 发布

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分类专栏： LightOJ algorithm 文章标签： DP

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本文介绍了一道名为DigitCount的问题解决思路与DP算法实现，该问题要求从给定的一组数字中找到所有符合条件的n位数，条件是任意相邻两个数字之间的差不超过2。

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题目链接：http://lightoj.com/volume_showproblem.php?problem=1122

1122 - Digit Count

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Given a set of digits S, and an integer n, you have to find how many n-digit integers are there, which contain digits that belong to S and the difference between any two adjacent digits is not more than two.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case contains two integers, m (1 ≤ m < 10) and n (1 ≤ n ≤ 10). The next line will contain m integers (from 1 to 9) separated by spaces. These integers form the set S as described above. These integers will be distinct and given in ascending order.

Output

For each case, print the case number and the number of valid n-digit integers in a single line.

Sample Input

Output for Sample Input

3 2

1 3 6

3 2

1 2 3

3 3

1 4 6

Case 1: 5

Case 2: 9

Case 3: 9

Note

For the first case the valid integers are

PROBLEM SETTER: JANE ALAM JAN

题意：从给出的m个数字，让你组成一个n位数，要求是任何两位之间差的绝对值不能大于2

解析：DP，dp[i][j] 是第i位为a[j] 的方案数，dp[i][j] += dp[i-1][jj], (abs(a[j] - a[jj]) <= 2

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<cmath>
#include<string>
#define N 1009
using namespace std;
const int INF = 0x3f3f3f3f;

int dp[N][N], a[N];

int main()
{
    int t, m, n, cnt = 0;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &m, &n);
        for(int i = 1; i <= m; i++) scanf("%d", &a[i]);
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= m; i++) dp[1][i] = 1;
        for(int i = 2; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                for(int jj = 1; jj <= m; jj++) if(abs(a[j] - a[jj]) <= 2) dp[i][j] += dp[i - 1][jj];
            }
        }
        int ans = 0;
        for(int i = 1; i <= m; i++) ans += dp[n][i];
        printf("Case %d: %d\n", ++cnt, ans);
    }
    return 0;
}