HDU 5375 Gray code（DP）

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本文探讨了灰码（格雷码）的概念及其在数字通信系统中的应用，通过实例展示了如何利用灰码来优化计分过程，实现更高效的数据传输与处理。文中详细介绍了灰码的特性、转换方法及在特定场景下的应用策略，为读者提供了深入理解灰码在实际工程中的实用技巧。

Gray code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 185 Accepted Submission(s): 105

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

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")
#define eps 1e-8
typedef long long ll;

using namespace std;
#define mod 1000000007

#define INF 0x3f3f3f3f
#define N 200005

int va[N];

char c[N];
int len;
int ans;
int dp[N][2];
int ca=0;

void DP()
{

   int i,j;
   if(c[1]=='1')
   {
       dp[1][1]=va[1];
       dp[1][0]=-INF;
   }
   else
    if(c[1]=='0')
   {
       dp[1][1]=-INF;
       dp[1][0]=0;
   }
   else
   {
       dp[1][0]=0;
       dp[1][1]=va[1];
   }

   for(i=2;i<=len;i++)
      if(c[i]=='1')
   {
       dp[i][1]=max(dp[i-1][1],dp[i-1][0]+va[i]);
       dp[i][0]=-INF;
   }
   else if(c[i]=='0')
   {
       dp[i][1]=-INF;
       dp[i][0]=max(dp[i-1][0],dp[i-1][1]+va[i]);
   }
   else
   {
       dp[i][1]=max(dp[i-1][1],dp[i-1][0]+va[i]);
       dp[i][0]=max(dp[i-1][1]+va[i],dp[i-1][0]);
   }

   printf("Case #%d: %d\n",++ca,max(dp[len][0],dp[len][1]));
}

int main()
{
    int i,j,t,ca=0;
    scanf("%d",&t);
    while(t--)
    {
      scanf("%s",c+1);
      len=strlen(c+1);
      for(i=1;i<=len;i++)
         scanf("%d",&va[i]);
      DP();
    }
   return 0;
}