UVALive - 4764 bing it——动态规划

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 N rounds 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 integer N (2 ≤ N ≤ 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

题意：给定一串数字序列 从前到后依次选择bing或者不bing 如果这张牌和bing了的一样 就加一分 999的话加三分 问最多可以得到多少分

思路：动态规划 用dp[i]记录前i张牌可以获得的最大分数 用num[x]记录bing了x的上一张牌是第几张 -1表示还没有被bing过

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#define max_ 100010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int n;
int num[max_],dp[max_];
int main(int argc, char const *argv[])
{
int i;
while(scanf("%d",&n)!=EOF&&n!=0)
{
memset(dp,0,sizeof(dp));
memset(num,-1,sizeof(num));
for(i=1;i<=n;i++)
{
int x;
scanf("%d",&x);
if(num[x]==-1)
dp[i]=dp[i-1];
else
{
if(x==999)
dp[i]=max(dp[i-1],dp[num[x]]+3);
else
dp[i]=max(dp[i-1],dp[num[x]]+1);
}
num[x]=i;
}
printf("%d\n",dp
);
}
return 0;
}