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,Gi=Bi^Bi+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=0;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?1:-1; 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?1:-1;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=0;i<b.size();i++)a[i]=b[i];} template<typename T> void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} const double PI = acos(-1.0); const int INF = 1e9 + 7; const double EPS = 1e-8; /* -------------------------------------------------------------------------------- */ const int maxn = 2e5 + 7; int dp[maxn][2], 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 = 0; cin >> T; while (T --) { scanf("%s", s); int n = strlen(s); for (int i = 0; i < n; i ++) scanf("%d", a + i); fillchar(dp, 0); for (int i = n - 2; i >= 0; i --) { if (s[i] == '0') { if (s[i + 1] == '0') dp[i][0] = dp[i + 1][0]; if (s[i + 1] == '1') dp[i][0] = dp[i + 1][1] + a[i + 1]; if (s[i + 1] == '?') dp[i][0] = max(dp[i + 1][0], dp[i + 1][1] + a[i + 1]); } if (s[i] == '1') { if (s[i + 1] == '0') dp[i][1] = dp[i + 1][0] + a[i + 1]; if (s[i + 1] == '1') dp[i][1] = dp[i + 1][1]; if (s[i + 1] == '?') dp[i][1] = max(dp[i + 1][0] + a[i + 1], dp[i + 1][1]); } if (s[i] == '?') { if (s[i + 1] == '0') { dp[i][0] = dp[i + 1][0]; dp[i][1] = dp[i + 1][0] + a[i + 1]; } if (s[i + 1] == '1') { dp[i][0] = dp[i + 1][1] + a[i + 1]; dp[i][1] = dp[i + 1][1]; } if (s[i + 1] == '?') { dp[i][0] = max(dp[i + 1][0], dp[i + 1][1] + a[i + 1]); dp[i][1] = max(dp[i + 1][0] + a[i + 1], dp[i + 1][1]); } } } int ans; if (s[0] == '0') ans = dp[0][0]; if (s[0] == '1') ans = dp[0][1] + a[0]; if (s[0] == '?') ans = max(dp[0][0], dp[0][1] + a[0]); printf("Case #%d: %d ", ++ cas, ans); } return 0; }