题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5375

题面：

Gray code

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

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

Sample Output

Case #1: 12
Case #2: 15

Hint

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

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


 

Source

2015 Multi-University Training Contest 7

解题：

类似2014区域赛的一道简单dp。首先要知道格雷码是怎么算的，格雷码就是当前这一位的二进制表示和其前一位的二进制表示相同时，取0否则取1.（认为最高位前面一位为0）。

状态转移方程为:

dp[i][0]=max(dp[i-1][0],dp[i-1][1]+a[i]);

dp[i][1]=max(dp[i-1][1],dp[i-1][0]+a[i]);

但因为某些位是给定的，所以只能算其中部分状态，加一些if判断即可，详见代码。如有问题，欢迎交流！！

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;
}

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