DP->UVALive 4764

最新推荐文章于 2018-09-29 00:45:45 发布

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最新推荐文章于 2018-09-29 00:45:45 发布

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分类专栏： ACM->DP

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本文介绍了一款名为Bingit的游戏，该游戏包含N轮，每轮会开启一种编号从1到100000的新卡片。玩家可以通过bing特定的卡片来获取分数，若连续两次出现相同的卡片则可获得1分，特殊卡片999连续出现可得3分。文章提供了求解最大得分的算法实现，并附带示例输入输出。

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Bing it

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status Practice UVALive 4764

Description

I guess most of you played cards on the trip to Harbin, but I'm sure you have never played the following card game. This card game has Nrounds and 100000 types of cards numbered from 1 to 100000. A new card will be opened when each round begins. You can ``bing" this new card. And if the card you last ``bing" is the same with this new one, you will get 1 point. You can "bing" only one card, but you can change it into a new one. For example, the order of the 4 cards is 1 3 4 3. You can ``bing" 1 in the first round and change it into 3 in the second round. You get no point in the third round, but get 1 point in the last round. Additionally, there is a special card 999. If you ``bing" it and it is opened again, you will get 3 point.

Given the order of N cards, tell me the maximum points you can get.

Input

The input file will contain multiple test cases. Each test case will consist of two lines. The first line of each test case contains one integerN(2 $\le$ N $\le$ 100000). The second line of each test case contains a sequence of n integers, indicating the order of cards. A single line with the number ``0" marks the end of input; do not process this case.

Output

For each input test case, print the maximum points you can get.

Sample Input

2 
1 1 
5 
1 999 3 3 999 
0

Sample Output

1 
3

/*This Code is Submitted by OUC_Fighting for Problem 2883 at 2013-10-06 11:20:28*/
#include <iostream>
#include <string.h>

using namespace std;

#define MAXN 100000 + 10

int temp[MAXN], dp[MAXN];
bool vis[MAXN];

void input()
{
    int n;

    while (cin >> n, n)
    {
        for (int i = 0; i < n; i++)
        {
            cin >> temp[i];
        }

        dp[temp[0]] = 0;

        int per = temp[0], mx = -1;

        memset(dp, 0, sizeof(dp));
        memset(vis, false, sizeof(vis));
        vis[per] = true;

        for (int i = 1; i < n; i++)
        {
            int next = temp[i];

            if (!vis[next])
            {
                dp[next] = dp[per];
                vis[next] = true;
            }
            else
            {
                if (next == 999)
                {
                    dp[next] = max(dp[per], dp[next] + 3);
                }
                else
                {
                    dp[next] = max(dp[per], dp[next] + 1);
                }
            }

            per = next;

            if (dp[next] > mx)
            {
                mx = dp[next];
            }
        }

        cout << mx << endl;
    }
}

int main()
{
    std::ios::sync_with_stdio(false);
    input();
    return 0;
}