递推DP HDOJ 5375 Gray code

 

题目传送门

 1 /*
 2     题意：给一个串，只能是0，1，?(0/1)。计算格雷码方法：当前值与前一个值异或，若为1，可以累加a[i],问最大累加值
 3     DP：dp[i][0/1]表示当前第i位选择0/1时的最大分数，那么分类讨论，情况太多，看代码，注意不可能的状态不要转移
 4 */
 5 /************************************************
 6 * Author        :Running_Time
 7 * Created Time  :2015-8-11 15:49:03
 8 * File Name     :G.cpp
 9  ************************************************/
10 
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <sstream>
15 #include <cstring>
16 #include <cmath>
17 #include <string>
18 #include <vector>
19 #include <queue>
20 #include <deque>
21 #include <stack>
22 #include <list>
23 #include <map>
24 #include <set>
25 #include <bitset>
26 #include <cstdlib>
27 #include <ctime>
28 using namespace std;
29 
30 #define lson l, mid, rt << 1
31 #define rson mid + 1, r, rt << 1 | 1
32 typedef long long ll;
33 const int MAXN = 2e5 + 10;
34 const int INF = 0x3f3f3f3f;
35 const int MOD = 1e9 + 7;
36 char s[MAXN];
37 int a[MAXN];
38 int dp[MAXN][2];
39 
40 int work(int n)    {
41     memset (dp, 0, sizeof (dp));
42     if (s[1] == '?')    dp[1][1] = a[1];
43     else    {
44         int t = s[1] - '0';
45         if (t == 1)    dp[1][1] = a[1];
46     }
47     for (int i=2; i<=n; ++i)    {
48         if (s[i] == '?')    {
49             if (s[i-1] == '?')    {
50                 dp[i][0] = max (dp[i-1][0], dp[i-1][1] + a[i]);
51                 dp[i][1] = max (dp[i-1][1], dp[i-1][0] + a[i]);
52             }
53             else    {
54                 int t = s[i-1] - '0';
55                 dp[i][1-t] = max (dp[i-1][1-t], dp[i-1][t] + a[i]);
56                 dp[i][t] = dp[i-1][t];
57             }
58         }
59         else    {
60             if (s[i-1] == '?')    {
61                 int t = s[i] - '0';
62                 dp[i][t] = max (dp[i-1][t], dp[i-1][1-t] + a[i]);
63             }
64             else    {
65                 int tt = s[i-1] - '0', t = s[i] - '0';
66                 dp[i][t] = dp[i-1][tt] + ((tt != t) ? a[i] : 0);
67             }
68         }
69     }
70 
71     if (s[n] == '?')    {
72         return max (dp[n][0], dp[n][1]);
73     }
74     else return dp[n][s[n]-'0'];
75 }
76 
77 int main(void)    {        //HDOJ 5375 Gray code
78     int T, cas = 0;    scanf ("%d", &T);
79     while (T--)    {
80         scanf ("%s", s + 1);
81         int len = strlen (s + 1);
82         for (int i=1; i<=len; ++i)    scanf ("%d", &a[i]);
83         printf ("Case #%d: %d\n", ++cas, work (len));
84     }
85 
86     return 0;
87 }

 

posted @ 2015-08-12 10:39  Running_Time  阅读(177)  评论(0)  编辑  收藏  举报