Time solution to a problem Acwing P288 rest

Analysis

First, assume the first hour of the day is not connected to the N-th day after hours, DP transfer miss compare this case

Consider before i hours a day, have a rest j hours, and the i-th hour whether at rest

Then there is the state transition equation:

dp[i][j][0]=max(dp[i-1][j][0],dp[i-1][j][1]);

dp[i][j][1]=max(dp[i-1][j-1][0],dp[i-1][j-1][1]+a[i]);

Is initialized to

Answer is

Connected to consider the N-th day and hour of the day after the first hour now

We found that the above transfer, the only thing taken into account only if the first one hour of rest to get physical

So we can initialize $U_1 D P [ . 1 ] [ . 1 ] [ . 1 ] = the U- . 1 , the same as above transfer equation$

Then the answer to

So far here already AC, but! ! ! If we get the submission POJ, you will find yourself crazy MLE (POJ frenzied Memory limit only 64M)

So we consider the use of an array of space-optimized rolling

dp[i&1][j][0]=max(dp[(i-1)&1][j][0],dp[(i-1)&1][j][1]);

dp[i&1][j][1]=max(dp[(i-1)&1][j-1][0],dp[(i-1)&1][j-1][1]+a[i]);

Because dp [I] [] [] with only dp [i-1] [] [] related, so long as the alternating array of dimension 0 and dimension 1, save only the last update dp array can greatly optimize the space

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define int long long 
 6 #define maxn 3830+10
 7 using namespace std;
 8 inline int read()
 9 {
10     int x=0;
11     bool f=1;
12     char c=getchar();
13     for(; !isdigit(c); c=getchar()) if(c=='-') f=0;
14     for(; isdigit(c); c=getchar()) x=(x<<3)+(x<<1)+c-'0';
15     if(f) return x;
16     return 0-x;
17 }
18 inline void write(int x)
19 {
20     if(x<0){putchar('-');x=-x;}
21     if(x>9)write(x/10);
22     putchar(x%10+'0');
23 }
24 int n,b,ans;
25 int u[maxn];
26 int dp1[2][maxn][2],dp2[2][maxn][2];
27 signed main()
28 {
29 //    freopen("naptime.in","r",stdin);
30 //    freopen("naptime.out","w",stdout);
31     n=read();b=read();
32     for(int i=1;i<=n;i++) u[i]=read();
33     memset(dp1,128,sizeof(dp1));
34     dp1[1][0][0]=dp1[1][1][1]=0;
35     memset(dp2,128,sizeof(dp2));
36     dp2[1][1][1]=u[1];
37     for(int i=2;i<=n;i++)
38     {
39         for(int j=0;j<=min(i,b);j++)
40         {
41             dp1[i&1][j][0]=max(dp1[(i-1)&1][j][0],dp1[(i-1)&1][j][1]);
42             if(j>=1) dp1[i&1][j][1]=max(dp1[(i-1)&1][j-1][0],dp1[(i-1)&1][j-1][1]+u[i]);
43             dp2[i&1][j][0]=max(dp2[(i-1)&1][j][0],dp2[(i-1)&1][j][1]);
44             if(j>=1) dp2[i&1][j][1]=max(dp2[(i-1)&1][j-1][0],dp2[(i-1)&1][j-1][1]+u[i]);
45         }
46     }
47     ans=max(dp2[n&1][b][1],max(dp1[n&1][b][0],dp1[n&1][b][1]));
48     write(ans);
49     return 0;
50 }

Please Gangster treatise~~(Anyway, I do not know what that means treatise)~~

Time solution to a problem Acwing P288 rest

Analysis

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