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Gray code

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.

2
00?0
1 2 4 8
????
1 2 4 8

Case #1: 12
Case #2: 15
Hint
https://en.wikipedia.org/wiki/Gray_code

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

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define INF 1000000000
using namespace std;
char s[200005];
int a[200005],dp[200005][2];
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int t,len,i,cnt=1;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        scanf("%s",s+1);
        len=strlen(s+1);
        for(i=1;i<=len;i++)
        {
            scanf("%d",&a[i]);
            getchar();
        }
        memset(dp,0,sizeof(dp));
        if(s[1]=='0')
            dp[1][0]=0;
        else if(s[1]=='1')
            dp[1][1]=a[1];
        else
        {
            dp[1][1]=a[1];
            dp[1][0]=0;
        }
        for(i=2;i<=len;i++)
        {
            if(s[i]=='0')
            {
                if(s[i-1]=='0')
                    dp[i][0]=dp[i-1][0];
                else if(s[i-1]=='1')
                    dp[i][0]=dp[i-1][1]+a[i];
                else
                    dp[i][0]=max(dp[i-1][0],dp[i-1][1]+a[i]);
            }
            else if(s[i]=='1')
            {
                if(s[i-1]=='0')
                    dp[i][1]=dp[i-1][0]+a[i];
                else if(s[i-1]=='1')
                    dp[i][1]=dp[i-1][1];
                else 
                  dp[i][1]=max(dp[i-1][0]+a[i],dp[i-1][1]);
            }
            else
            {
                if(s[i-1]=='0')
                {
                    dp[i][1]=dp[i-1][0]+a[i];
                    dp[i][0]=dp[i-1][0];
                }
                else if(s[i-1]=='1')
                {
                    dp[i][1]=dp[i-1][1];
                    dp[i][0]=dp[i-1][1]+a[i];
                }
                else 
                {
                    dp[i][1]=max(dp[i-1][0]+a[i],dp[i-1][1]);
                    dp[i][0]=max(dp[i-1][0],dp[i-1][1]+a[i]);
                }
            }
        }
        printf("Case #%d: ",cnt++);
        printf("%d\n",max(dp[len][0],dp[len][1]));
    }
    return 0;
}

HDU 5375 Gray code （简单dp）

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原文地址：http://blog.csdn.net/david_jett/article/details/47606187