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Gray code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 684 Accepted Submission(s): 402


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

2
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
 

題意：給你一串二進制代碼。？既能夠表示1也能夠表示0。

將其轉換成格雷碼，每一位都有相應的數值，問將全部格雷碼為1的數值相加，求最大值。

題解：首先我們要知道二進制碼怎麽轉換為格雷碼。將所需轉換的數右移一位再與原數異或就可以。這裏我們能夠發現當前位置的取值至於下一位有關，所以我們想能夠用dp解決此問題。

參考代碼：

#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) 解題報告