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HDOJ 5375 Gray code DP

編輯：關於C++

標准格雷碼的性質：二進制a1 a2 ... an，對應的格雷碼為a1 （a1 xor a2） ... （an-1 xor an）題目就可以轉為O(n)的dp dp[i][j]表示二進制第i位為j時前i位對應最大分數.

Gray code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 298 Accepted Submission(s): 163

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

Source 2015 Multi-University Training Contest 7

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月11日 星期二 20時56分37秒
File Name     :HDOJ5375.cpp
************************************************ */

#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

const int maxn=200200;

int n;
char str[maxn];
int a[maxn];
int dp[maxn][2];

int main()
{
    //freopen(in.txt,r,stdin);
    //freopen(out.txt,w,stdout);

    int cas=1,T_T;
    scanf(%d,&T_T);
    while(T_T--)
    {
        scanf(%s,str);
        n=strlen(str);
        for(int i=0;i 0
                if(str[i-1]=='?'||str[i-1]=='1') dp[i][0]=max(dp[i][0],dp[i-1][1]+a[i]);
                if(str[i-1]=='?'||str[i-1]=='0') dp[i][0]=max(dp[i][0],dp[i-1][0]);
                /// ? --> 1
                if(str[i-1]=='?'||str[i-1]=='0') dp[i][1]=max(dp[i][1],dp[i-1][0]+a[i]);
                if(str[i-1]=='?'||str[i-1]=='1') dp[i][1]=max(dp[i][1],dp[i-1][1]);
            }
        }

		int ans=max(dp[n-1][0],dp[n-1][1]);

        printf(Case #%d: %d
,cas++,ans);
    }
    
    return 0;
}