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Gray code

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.

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

Case #1: 12
Case #2: 15
Hint
https://en.wikipedia.org/wiki/Gray_code

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

题意：给你一串二进制代码，？既可以表示1也可以表示0。将其转换成格雷码，每一位都有对应的数值，问将所有格雷码为1的数值相加，求最大值。

题解：首先我们要知道二进制码怎么转换为格雷码。将所需转换的数右移一位再与原数异或即可。这里我们可以发现当前位置的取值至于下一位有关，所以我们想可以用dp解决此问题。

参考代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
#define max(a,b) a>b?a:b
using namespace std;
const int N = 200005;
const int INF = 999999999;
char s[N];
int dp[N][2], num[N];

int main() {
    int T, Case = 1;
    scanf("%d", &T);
    while (T--) {
        getchar();
        gets(s);    
        int len = strlen(s);
        for (int i = 0; i < len; i++) {
            scanf("%d", &num[i]);   
        }
        if (s[0] == '1') {
            dp[0][1] = num[0];      
            dp[0][0] = -INF;
        } else if (s[0] == '0') {
            dp[0][1] = -INF;
            dp[0][0] = 0;
        } else {
            dp[0][1] = num[0];
            dp[0][0] = 0;
        }
        for (int i = 1; i < len; i++) {
            if (s[i] == '0') {
                dp[i][0] = max(dp[i - 1][1] + num[i], dp[i - 1][0]);
                dp[i][1] = -INF;    
            } else if (s[i] == '1') {
                dp[i][1] = max(dp[i - 1][0] + num[i], dp[i - 1][1]);
                dp[i][0] = -INF;
            } else {
                dp[i][1] = max(dp[i - 1][0] + num[i], dp[i - 1][1]);
                dp[i][0] = max(dp[i - 1][1] + num[i], dp[i - 1][0]);
            }
        }
        printf("Case #%d: ", Case++);   
        printf("%d\n", max(dp[len - 1][1], dp[len - 1][0]));
    }
    return 0;
}

hdu 5375 - Gray code（dp) 解题报告

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原文地址：http://blog.csdn.net/luwhere/article/details/47663437