HDU 5375

时间：2015-08-13 12:23:52 阅读：146 评论：0 收藏：0 [点我收藏+]

Gray code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 627 Accepted Submission(s): 370

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.

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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

Source

2015 Multi-University Training Contest 7

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wange2014 | We have carefully selected several similar problems for you: 5379 5378 5377 5376 5374

第一个例子：

00?0

此时默认左边为0 即000?0。

所以可以得到0^0=0,0^0=0。?代表可以为0或者1.如果为1的话，则0^1=1,1^0=1 得到0011 就算出0*1+0*2+1*4+1*8 如果为0的话那么全为0 0000 就算出0.

题意显然是问你？填多少时，值是最大的。这个序列的长度为20W直接暴力肯定跪。

容易想到动态规划，，以dp[i][1] 代表在i这个位置为1的情况，dp[i][0]代表在i这个位置为0的情况，我们枚举每一个i的前面的为0，1，？的三种情况进行讨论。遇到?在进行两种情况的讨论，不断的更新答案，就可以了。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
typedef long long LL;
const int MAXN = 210000;//点数的最大值
const int MAXM = 604000;//边数的最大值
const LL INF = 1152921504;
const LL mod= 258280327;
int num[MAXN],b[MAXN],a[MAXN];
char s[MAXN];
int dp[MAXN][2];
int main()
{
    int t,cas=1;
    cin>>t;
    while(t--)
    {
        scanf("%s",s+1);
        int n=strlen(s+1);
        memset(dp,0,sizeof(dp));
        a[0]=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&num[i]);
            if(s[i]=='?')
                a[i]=2;
            else a[i]=s[i]-'0';
        }
        for(int i=1; i<=n; i++)
        {
            if(a[i]==2)
            {
                if(a[i-1]==2)
                {
                    dp[i][1]=max(dp[i-1][0]+num[i],dp[i][1]);
                    dp[i][1]=max(dp[i-1][1],dp[i][1]);
                }
                else if(a[i-1]==0)
                    dp[i][1]=max(dp[i-1][0]+num[i],dp[i][1]);
                else
                    dp[i][1]=max(dp[i-1][1],dp[i][1]);
                if(a[i-1]==2)
                {
                    dp[i][0]=max(dp[i-1][1]+num[i],dp[i][0]);
                    dp[i][0]=max(dp[i-1][0],dp[i][0]);
                }
                else if(a[i-1]==1)
                {
                    dp[i][0]=max(dp[i-1][1]+num[i],dp[i][0]);
                }
                else
                {
                    dp[i][0]=max(dp[i-1][0],dp[i][0]);
                }
            }
            else if(a[i]==1)
            {

                if(a[i-1]==2)
                {
                    dp[i][1]=max(dp[i-1][0]+num[i],dp[i][1]);
                    dp[i][1]=max(dp[i-1][1],dp[i][1]);
                }
                else if(a[i-1]==0)
                    dp[i][1]=max(dp[i-1][0]+num[i],dp[i][1]);
                else
                    dp[i][1]=max(dp[i-1][1],dp[i][1]);
            }
            else if(a[i]==0)
            {
                if(a[i-1]==2)
                {
                    dp[i][0]=max(dp[i-1][1]+num[i],dp[i][0]);
                    dp[i][0]=max(dp[i-1][0],dp[i][0]);
                }
                else if(a[i-1]==1)
                {
                    dp[i][0]=max(dp[i-1][1]+num[i],dp[i][0]);

                }
                else
                {
                    dp[i][0]=max(dp[i-1][0],dp[i][0]);
                }
            }
        }
        printf("Case #%d: %d\n",cas++,max(dp[n][1],dp[n][0]));
    }
    return 0;
}
/*
2
??110
1 2 3 4 5
*/

HDU 5375

标签：acm算法

原文地址：http://blog.csdn.net/sky_miange/article/details/47606487

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