[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,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;
}
原文地址:https://www.cnblogs.com/jklongint/p/4724144.html