hdu 5375 Gray code_binocular gray code-优快云博客

本文探讨了灰码（格雷码）的应用及其在特定条件下的优化策略，详细阐述了如何通过选择最优路径来最大化得分的过程。通过实例分析，展示了在给定的字符串和分数序列下，如何计算出最高得分，提供了深入理解灰码在实际问题解决中的实用方法。

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Gray code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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

无力吐槽，先找出"?"段ab,如果ab两边的值相等且ab长为奇数，那么a ，b+1位置上都可以异或1，如果ab位置上的数不同，而且ab长度为偶数，也可以。

如果不符合这些，那么先把a+1到b+1的位置都按可以异或成1算，然后再减去一个权值最小的位置。就是从两边贪心，是0则1，是1则0。

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<climits>
#include<queue>
#include<vector>
#include<map>
#include<sstream>
#include<set>
#include<stack>
#include<utility>
#pragma comment(linker, "/STACK:102400000,102400000")
#define PI 3.1415926535897932384626
#define eps 1e-10
#define sqr(x) ((x)*(x))
#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)
#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)
#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)
#define  lson   num<<1,le,mid
#define rson    num<<1|1,mid+1,ri
#define MID   int mid=(le+ri)>>1
#define zero(x)((x>0? x:-x)<1e-15)

using namespace std;
const int INF =0x3f3f3f3f;
const int maxn= 200000   ;
//const int maxm=    ;
//const int INF=    ;
typedef long long ll;
const ll inf =1000000000000000;//1e15;
//ifstream fin("input.txt");
//ofstream fout("output.txt");
//fin.close();
//fout.close();
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);
char s[maxn+20];
int a[maxn+20];
int n;
int cnt,kase=0;
pair<  int,int>  se[maxn+20];

void check()
{
    for(int i=1;i<=cnt;i++)
    {
        cout<<se[i].first<<" "<<se[i].second<<endl;
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",s+1);
        n=strlen(s+1);
        s[0]='0';
       int ans=0;
        bool has=0;cnt=0;int last;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);

            if(s[i]!='?')
            {
              if( !has )  {
//                  cout<<"hasadd :"<<i<<endl;
                    ans+= ((s[i-1]-'0')^ (s[i]-'0') )*a[i];continue;}
               se[++cnt]=make_pair(last,i-1   );
               has=0;

            }
            else
            {
                if(has)  continue;

                has=1;last=i ;
            }
        }
//        check();
         printf("Case #%d: ",++kase);

        if( s[n]=='?' )  se[++cnt]=make_pair(last,n);

        for(int i=1;i<=cnt;i++)
        {
            pair<int,int> & now=se[i];
            if( se[i].second==n )
            {

                  for(int j=now.first;j<=now.second;j++)
                   {
                       ans+=  a[j];
                   }




            }
            else
            {
                  int len=now.second-now.first+1;
               if(s[now.first-1]==s[now.second+1]&& len%2==1 || s[now.first-1]!=s[now.second+1]&&len%2==0)
               {
                   for(int j=now.first;j<=now.second+1;j++)
                   {
                       ans+=  a[j];
                   }
               }
               else
               {


                int len=now.second-now.first+1;
                int mini=INF;
                for(int j=now.first;j<=now.second+1;j++)
                {
                    mini=min(a[j],mini);
                    ans+=a[j];
                }

                ans-=mini;
               }
            }


        }
        printf("%d\n",ans);


    }


    return 0;
}/*
00?0????
1 2 4 8 1 2 4 8

*/