HDU-5375 Gray code

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最新推荐文章于 2025-08-24 15:22:07 发布

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原文链接：http://www.cnblogs.com/cancangood/p/4723430.html

本文介绍了一种解决特定灰码问题的方法，通过算法确定给定包含不确定位的二进制字符串所能获得的最大积分。该问题涉及到灰码的概念及其在数字通信中的应用。

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Gray code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 422 Accepted Submission(s): 242

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

Source

2015 Multi-University Training Contest 7

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <malloc.h>
#define Max(a,b) (a>b?a:b)
#define Min(a,b) (a<b?a:b)
#define MAX 999999999
#define LL long long
#define M 6666666
using namespace std;
char s[200010];
int d[200010][2];
int main()
{
    int t,n,m,i,j,k,l,cas=1;
    scanf("%d",&t);
    while(t--&&scanf("%s",s+1))
    {
        s[0]='0';
        memset(d,-1,sizeof(d));
        d[0][0]=0;
        l=strlen(s);
        for(i=1;i<l;i++)
        {
            scanf("%d",&k);
            if(s[i]=='?')
            {
                if(d[i-1][0]>-1)
                {
                    d[i][1]=Max(d[i][1],d[i-1][0]+k);
                    d[i][0]=Max(d[i][0],d[i-1][0]);
                }
                if(d[i-1][1]>-1)
                {
                    d[i][1]=Max(d[i][1],d[i-1][1]);
                    d[i][0]=Max(d[i][0],d[i-1][1]+k);
                }
            }else if(s[i]=='1')
            {
                if(d[i-1][0]>-1)
                {
                    d[i][1]=Max(d[i][1],d[i-1][0]+k);
                }
                if(d[i-1][1]>-1)
                {
                    d[i][1]=Max(d[i][1],d[i-1][1]);
                }
            }else if(s[i]=='0')
            {
                if(d[i-1][0]>-1)
                {
                    d[i][0]=Max(d[i][0],d[i-1][0]);
                }
                if(d[i-1][1]>-1)
                {
                    d[i][0]=Max(d[i][0],d[i-1][1]+k);
                }
            }
        }
        printf("Case #%d: %d\n",cas++,Max(d[l-1][0],d[l-1][1]));
    }
    return 0;
}

转载于:https://www.cnblogs.com/cancangood/p/4723430.html