[hdu5375 Gray code]DP

题意：给一个二进制码，其中有一些位上为'?'，对每个问号确定是'0'还是'1'，最后以它对应的格雷码来取数，第i位为1则取第i个数，求取得的数的和的最大值。

思路：二进制码B转换成格雷码G的方法是，G_i=B_i^B_i+1，Gn=Bn。所以第i位如果为'?'，那么选'1'还是'0'只会影响邻位，于是用dp即可解决。

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

//#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
//#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
template<typename T>
void V2A(T a[],const vector<T>&b){for(int i=;i<b.size();i++)a[i]=b[i];}
template<typename T>
void A2V(vector<T>&a,const T b[]){for(int i=;i<a.size();i++)a[i]=b[i];}

const double PI = acos(-1.0);
const int INF = 1e9 + ;
const double EPS = 1e-8;

/* -------------------------------------------------------------------------------- */

const int maxn = 2e5 + ;

int dp[maxn][], a[maxn];
char s[maxn];
int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int T, cas = ;
    cin >> T;
    while (T --) {
        scanf("%s", s);
        int n = strlen(s);
        for (int i = ; i < n; i ++) scanf("%d", a + i);
        fillchar(dp, );
        for (int i = n - ; i >= ; i --) {
            if (s[i] == '0') {
                if (s[i + ] == '0') dp[i][] = dp[i + ][];
                if (s[i + ] == '1') dp[i][] = dp[i + ][] + a[i + ];
                if (s[i + ] == '?') dp[i][] = max(dp[i + ][], dp[i + ][] + a[i + ]);
            }
            if (s[i] == '1') {
                if (s[i + ] == '0') dp[i][] = dp[i + ][] + a[i + ];
                if (s[i + ] == '1') dp[i][] = dp[i + ][];
                if (s[i + ] == '?') dp[i][] = max(dp[i + ][] + a[i + ], dp[i + ][]);
            }
            if (s[i] == '?') {
                if (s[i + ] == '0') {
                    dp[i][] = dp[i + ][];
                    dp[i][] = dp[i + ][] + a[i + ];
                }
                if (s[i + ] == '1') {
                    dp[i][] = dp[i + ][] + a[i + ];
                    dp[i][] = dp[i + ][];
                }
                if (s[i + ] == '?') {
                    dp[i][] = max(dp[i + ][], dp[i + ][] + a[i + ]);
                    dp[i][] = max(dp[i + ][] + a[i + ], dp[i + ][]);
                }
            }
        }
        int ans;
        if (s[] == '0') ans = dp[][];
        if (s[] == '1') ans = dp[][] + a[];
        if (s[] == '?') ans = max(dp[][], dp[][] + a[]);
        printf("Case #%d: %d\n", ++ cas, ans);
    }
    return ;
}