Description
这里有一个n*m的矩阵，请你选出其中k个子矩阵，使得这个k个子矩阵分值之和最大。注意：选出的k个子矩阵不能相互重叠。
Input
第一行为n,m,k（1≤n≤100,1≤m≤2,1≤k≤10），接下来n行描述矩阵每行中的每个元素的分值(每个元素的分值的绝对值不超过32767)。
Output
只有一行为k个子矩阵分值之和最大为多少。
Sample Input
3 2 2
1 -3
2 3
-2 3
Sample Output
9
因为m≤2，所以就成了sbDP题，分情况DP即可
1 var
2 f1:array[0..100,0..10]of longint;
3 s1:array[0..100]of longint;
4 f2:array[0..100,0..100,0..10]of longint;
5 s2:array[0..100,1..2]of longint;
6 n,m,k:longint;
7
8 procedure up(var x:longint;y:longint);
9 begin
10 if x<y then x:=y;
11 end;
12
13 procedure work1;
14 var
15 i,j,l:longint;
16 begin
17 fillchar(f1,sizeof(f1),200);
18 for i:=1 to n do
19 begin
20 read(s1[i]);
21 inc(s1[i],s1[i-1]);
22 end;
23 for i:=0 to n do
24 f1[i,0]:=0;
25 for l:=1 to k do
26 for i:=1 to n do
27 begin
28 up(f1[i,l],f1[i-1,l]);
29 for j:=1 to i do
30 up(f1[i,l],f1[j-1,l-1]+s1[i]-s1[j-1]);
31 end;
32 writeln(f1[n,k]);
33 end;
34
35 procedure work2;
36 var
37 i,j,l,r:longint;
38 begin
39 for i:=1 to n do
40 for j:=1 to m do
41 begin
42 read(s2[i,j]);
43 inc(s2[i,j],s2[i-1,j]);
44 end;
45 fillchar(f2,sizeof(f2),200);
46 for i:=0 to n do
47 for j:=0 to n do
48 f2[i,j,0]:=0;
49 for l:=1 to k do
50 for i:=1 to n do
51 for j:=1 to n do
52 begin
53 up(f2[i,j,l],f2[i-1,j,l]);
54 up(f2[i,j,l],f2[i,j-1,l]);
55 for r:=1 to i do
56 up(f2[i,j,l],f2[r-1,j,l-1]+s2[i,1]-s2[r-1,1]);
57 for r:=1 to j do
58 up(f2[i,j,l],f2[i,r-1,l-1]+s2[j,2]-s2[r-1,2]);
59 if i=j then
60 for r:=1 to i do
61 up(f2[i,j,l],f2[r-1,r-1,l-1]+s2[j,2]-s2[r-1,2]+s2[i,1]-s2[r-1,1]);
62 end;
63 writeln(f2[n,n,k]);
64 end;
65
66 begin
67 read(n,m,k);
68 if m=1 then work1
69 else work2;
70 end.
1084: [SCOI2005]最大子矩阵 - BZOJ,布布扣,bubuko.com
时间： 2024-12-29 11:29:04