HDU 5375 Gray code

最新推荐文章于 2019-02-27 14:32:00 发布

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最新推荐文章于 2019-02-27 14:32:00 发布

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本文介绍了一种基于动态规划的方法来解决灰码最大值问题，即如何从包含不确定位的二进制数中找到能够获得最大积分的灰码表示。

Gray code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 904 Accepted Submission(s): 510

Problem Description

The reflected binary code, also known as Gray code after Frank Gray, is a binary numeral system where two successive values differ in only onebit (binary digit). The reflected binary code was originally designed to prevent spurious output from electromechanical switches. Today, Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.

Now , you are given a binary number of length n including ‘0’ , ’1’ and ‘?’(? means that you can use either 0 or 1 to fill this position) and n integers(a1,a2,….,an) . A certain binary number corresponds to a gray code only. If the ith bit of this gray code is 1,you can get the point ai.
Can you tell me how many points you can get at most?

For instance, the binary number “00?0” may be “0000” or “0010”,and the corresponding gray code are “0000” or “0011”.You can choose “0000” getting nothing or “0011” getting the point a3 and a4.

Input

The first line of the input contains the number of test cases T.

Each test case begins with string with ‘0’,’1’ and ‘?’.

The next line contains n (1<=n<=200000) integers (n is the length of the string).

a1 a2 a3 … an (1<=ai<=1000)

Output

For each test case, output “Case #x: ans”, in which x is the case number counted from one,’ans’ is the points you can get at most

Sample Input

Sample Output

  
   Case #1: 12
Case #2: 15

   
    
     Hint
    
https://en.wikipedia.org/wiki/Gray_code

http://baike.baidu.com/view/358724.htm

Author

UESTC

Source

2015 Multi-University Training Contest 7

Recommend

输入是二进制的，能转换成格雷码，若格雷码对应位是1，那么就能取到这一位对应的值，求输入情况中，能取得最大值。这是一个DP题，DP[i][j]表示第I位取1或者0，所以，dp[i][0] = dp[i-1][1]+num[i],应为如果二进制中，I和I-1不相等，那么格雷码这一位就是1，就要取值。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long

ll num[200010];
ll dp[200010][2];
int main()
{
	ll T;
	cin >> T;
	ll g = 1;
	while(T--)
	{
		printf("Case #%lld: ",g++);
		string s;
		cin >> s;
		for(int i=0;i<s.length();i++)
		scanf("%lld",&num[i]);
		memset(dp,0,sizeof(dp));
		if(s[0]=='1'||s[0]=='?')dp[0][1] = num[0];
		for(int i=1;i<s.length();i++)
		{
			if (s[i - 1] == '?')
			{
				if (s[i] == '?')
				{
					dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + num[i]);
					dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + num[i]);
				}
				else if (s[i] == '0')
					dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + num[i]);
				else
					dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + num[i]);
			}
			else if (s[i - 1] == '0')
			{
				if (s[i] == '?')
				{
					dp[i][0] = dp[i - 1][0];
					dp[i][1] = dp[i - 1][0] + num[i];
				}
				else if (s[i] == '0')
					dp[i][0] = dp[i - 1][0];
				else
					dp[i][1] = dp[i - 1][0] + num[i];
			}
			else
			{
				if (s[i] == '?')
				{
					dp[i][0] = dp[i - 1][1]+num[i];
					dp[i][1] = dp[i - 1][1] ;
				}
				else if (s[i] == '0')
					dp[i][0] = dp[i - 1][1]+num[i];
				else
					dp[i][1] = dp[i - 1][1];
			}
		}
		cout<<max(dp[s.length()-1][0],dp[s.length()-1][1])<<endl;
	}
	return 0;
}