Gray code（简单dp水题）

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Gray code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1403 Accepted Submission(s): 749

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

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

简单dp。就是算格雷码。当前一位和上一位不一样就是1，一样就是0.

状态判断下转移就好了。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=200010;
int dp[maxn][3];
int money[maxn];
char s[maxn];
int main()
{
    int t;
    int cas=0;
    scanf("%d",&t);
    while(t--){
        cas++;
        memset(dp,0,sizeof(dp));
        scanf("%s",s);
        int len=strlen(s);
        for(int i=0;i<len;i++){
            scanf("%d",&money[i]);
        }
        int ans=0;
        if(s[0]=='0'){
            dp[0][0]=0;
            dp[0][1]=0;
        }
        else {
            dp[0][1]=money[0];
            dp[0][0]=0;
        }
        for(int i=1;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]+money[i];
                }
                else{
                    dp[i][0]=max(dp[i-1][1]+money[i],dp[i-1][0]);
                }
            }
            else if(s[i]=='1'){
                if(s[i-1]=='0'){
                    dp[i][1]=dp[i-1][0]+money[i];
                }
                else if(s[i-1]=='1'){
                    dp[i][1]=dp[i-1][1];
                }
                else{
                    dp[i][1]=max(dp[i-1][0]+money[i],dp[i-1][1]);
                }
            }
            else{
                if(s[i-1]=='0'){
                    dp[i][0]=dp[i-1][0];
                    dp[i][1]=dp[i-1][0]+money[i];
                }
                else if(s[i-1]=='1'){
                    dp[i][1]=dp[i-1][1];
                    dp[i][0]=dp[i-1][1]+money[i];
                }
                else{
                    dp[i][1]=max(dp[i-1][0]+money[i],dp[i-1][1]);
                    dp[i][0]=max(dp[i-1][0],dp[i-1][1]+money[i]);
                }
            }
        }
        if(s[len-1]=='0'){
            printf("Case #%d: %d\n",cas,dp[len-1][0]);
        }
        else if(s[len-1]=='1'){
            printf("Case #%d: %d\n",cas,dp[len-1][1]);
        }
        else{
            int haha=max(dp[len-1][1],dp[len-1][0]);
            printf("Case #%d: %d\n",cas,haha);
        }
        
    }
    return 0;
}

Author

UESTC