Luo Gu P1854 flower shop window display solution to a problem
Analysis
Given a matrix f * v
Required to travel from a first f-th row, each row a number is removed,
The number of columns and number of rows to be taken will be taken on a line of a large number, it can be removed seeking maximum
Note that the numbers in each row is removed must be larger than the number of columns of the row is the row number, etc.
Because they must leave enough to the front of the flower vase
Similarly each line can assume the maximum number must be less than or equal vase v- (f- number of the trekking)
Thus, we can easily draw the state transition equation
dp[i][j]=max(dp[i-1][k])+d[i][j](k<j)dp[i][j]=max(dp[i−1][k])+d[i][j](k<j)
Where DP [i] [j] D P [ i ] [ j ] indicates the maximum value that can be acquired from the first to travel to and remove the i-th row j-th row number
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stack>
6 #define int long long
7 #define maxn 100+10
8 #define INF 9223372036854775807
9 using namespace std;
10 inline int read()
11 {
12 int x=0;
13 bool f=1;
14 char c=getchar();
15 for(; !isdigit(c); c=getchar()) if(c=='-') f=0;
16 for(; isdigit(c); c=getchar()) x=(x<<3)+(x<<1)+c-'0';
17 if(f) return x;
18 return 0-x;
19 }
20 inline void write(int x)
21 {
22 if(x<0){putchar('-');x=-x;}
23 if(x>9)write(x/10);
24 putchar(x%10+'0');
25 }
26 int f,v;
27 int map[maxn][maxn];
28 int dp[maxn][maxn];
29 int path[maxn][maxn];
30 stack<int> s;
31 inline void print(int num,int x)
32 {
33 if(num==0) return;
34 s.push(x);
35 print(num-1,path[num][x]);
36 }
37 signed main()
38 {
39 freopen("flower.in","r",stdin);
40 freopen("flower.out","w",stdout);
41 memset(dp,128,sizeof(dp));
42 f=read();v=read();
43 for(int i=1;i<=f;i++)
44 for(int j=1;j<=v;j++)
45 map[i][j]=read();
46 for(int i=1;i<=v;i++) dp[1][i]=map[1][i];
47 for(int i=2;i<=f;i++)
48 for(int j=i;j<=v-(f-i);j++)
49 for(int k=1;k<j;k++)
50 {
51 if(dp[i-1][k]+map[i][j]>dp[i][j])
52 {
53 dp[i][j]=dp[i-1][k]+map[i][j];
54 path[i][j]=k;
55 }
56 }
57 int ans=-INF,fnum=0;
58 for(int i=1;i<=v;i++)
59 if(dp[f][i]>ans)
60 {
61 ans=dp[f][i];
62 fnum=i;
63 }
64 write(ans);
65 printf("\n");
66 print(f,fnum);
67 while(!s.empty())
68 {
69 write(s.top());
70 printf(" ");
71 s.pop();
72 }
73 return 0;
74 }
Please Gangster treatise(Anyway, I do not know what that means treatise)
Analysis
Given a matrix f * v
Required to travel from a first f-th row, each row a number is removed,
The number of columns and number of rows to be taken will be taken on a line of a large number, it can be removed seeking maximum
Note that the numbers in each row is removed must be larger than the number of columns of the row is the row number, etc.
Because they must leave enough to the front of the flower vase
Similarly each line can assume the maximum number must be less than or equal vase v- (f- number of the trekking)
Thus, we can easily draw the state transition equation
dp[i][j]=max(dp[i-1][k])+d[i][j](k<j)dp[i][j]=max(dp[i−1][k])+d[i][j](k<j)
Where DP [i] [j] D P [ i ] [ j ] indicates the maximum value that can be acquired from the first to travel to and remove the i-th row j-th row number
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stack>
6 #define int long long
7 #define maxn 100+10
8 #define INF 9223372036854775807
9 using namespace std;
10 inline int read()
11 {
12 int x=0;
13 bool f=1;
14 char c=getchar();
15 for(; !isdigit(c); c=getchar()) if(c=='-') f=0;
16 for(; isdigit(c); c=getchar()) x=(x<<3)+(x<<1)+c-'0';
17 if(f) return x;
18 return 0-x;
19 }
20 inline void write(int x)
21 {
22 if(x<0){putchar('-');x=-x;}
23 if(x>9)write(x/10);
24 putchar(x%10+'0');
25 }
26 int f,v;
27 int map[maxn][maxn];
28 int dp[maxn][maxn];
29 int path[maxn][maxn];
30 stack<int> s;
31 inline void print(int num,int x)
32 {
33 if(num==0) return;
34 s.push(x);
35 print(num-1,path[num][x]);
36 }
37 signed main()
38 {
39 freopen("flower.in","r",stdin);
40 freopen("flower.out","w",stdout);
41 memset(dp,128,sizeof(dp));
42 f=read();v=read();
43 for(int i=1;i<=f;i++)
44 for(int j=1;j<=v;j++)
45 map[i][j]=read();
46 for(int i=1;i<=v;i++) dp[1][i]=map[1][i];
47 for(int i=2;i<=f;i++)
48 for(int j=i;j<=v-(f-i);j++)
49 for(int k=1;k<j;k++)
50 {
51 if(dp[i-1][k]+map[i][j]>dp[i][j])
52 {
53 dp[i][j]=dp[i-1][k]+map[i][j];
54 path[i][j]=k;
55 }
56 }
57 int ans=-INF,fnum=0;
58 for(int i=1;i<=v;i++)
59 if(dp[f][i]>ans)
60 {
61 ans=dp[f][i];
62 fnum=i;
63 }
64 write(ans);
65 printf("\n");
66 print(f,fnum);
67 while(!s.empty())
68 {
69 write(s.top());
70 printf(" ");
71 s.pop();
72 }
73 return 0;
74 }
Please Gangster treatise(Anyway, I do not know what that means treatise)