HDU5375——DP——Gray code

 


格雷值与二进制转化公式,G[i] = B[i]^B[i-1],二进制前面补0
dp[i][j]定义为当前是第i个格雷值,值是j的和
* Author        :powatr
* Created Time  :2015-8-11 13:19:46
* File Name     :1007.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 2e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;

char s[MAXN];
int ss[MAXN][2];
int dp[MAXN][2];
int a[MAXN];
int main(){
    int T;
    scanf("%d", &T);
    for(int cas = 1; cas <= T; cas++){
        memset(a, 0, sizeof(a));
        memset(s, 0, sizeof(s));
        memset(dp, 0, sizeof(dp));
        scanf("%s", s+1);
        int len = strlen(s+1);
        for(int i = 1; i <= len; i++)
            scanf("%d", &a[i]);
        if(s[1] == '1') dp[1][1] = a[1];
        else if(s[1] == '0') dp[1][0] = 0;
        else if(s[1] == '?') {
           dp[1][1] = a[1], dp[1][0] = 0;
        }
        for(int i = 2; i <= len; i++){
            if(s[i] == '1'){
                if(s[i-1] == '0')
                    dp[i][1] = dp[i-1][0] + a[i];
                else if(s[i-1] == '1') dp[i][1] = dp[i-1][1];
                else if(s[i-1] == '?'){
                    dp[i][1] = max(dp[i-1][1], dp[i-1][0] + a[i]);
                }
            }
            else if(s[i] == '0'){
                if(s[i-1] == '1')
                    dp[i][0] = dp[i-1][1] + a[i];
                else if(s[i-1] == '0')  dp[i][0] = dp[i-1][0];
                else if(s[i-1] == '?'){
                    dp[i][0] = max(dp[i-1][1] + a[i], dp[i-1][0]);
                }
            }
            else if(s[i] == '?'){
                if(s[i-1] == '1'){
                    dp[i][0] = dp[i-1][1] + a[i];
                    dp[i][1] = dp[i-1][1] ;
                }
                else if(s[i-1] == '0'){
                    dp[i][0] = dp[i-1][0];
                    dp[i][1] = dp[i-1][0] + a[i];
                }
                else if(s[i-1] == '?'){
                    dp[i][0] = max(dp[i-1][0], dp[i-1][1]+a[i]);
                    dp[i][1] = max(dp[i-1][1], dp[i-1][0]+a[i]);
                }
            }
        }
        printf("Case #%d: %d\n",cas, max(dp[len][1], dp[len][0]));
    }
    return 0;
}

  

posted @ 2015-08-12 11:17  Painting、时光  阅读(152)  评论(0编辑  收藏  举报