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。
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。
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)。
只有一行为k个子矩阵分值之和最大为多少。
1 program bzoj1084; 2 3 {$mode objfpc}{$H+} 4 5 uses 6 Classes, SysUtils 7 { you can add units after this }; 8 9 var 10 i,j,k,l,m,n,tot,ii,jj:longint; 11 b:array [0..101,0..3] of longint; 12 a:array [0..101,0..4,0..10] of longint; 13 begin 14 readln(n,m,k); 15 for i:=1 to n do for j:=1 to m do read(b[i,j]); 16 if m=1 then 17 begin 18 for i:=0 to k do 19 begin 20 a[0,1,i]:=-maxlongint div 10; 21 a[0,0,i]:=-maxlongint div 10; 22 end; 23 a[0,0,0]:=0; 24 for i:=1 to n do 25 for j:=0 to k do 26 begin 27 if a[i-1,1,j]>a[i-1,0,j] then a[i,0,j]:=a[i-1,1,j] else a[i,0,j]:=a[i-1,0,j]; 28 if j=0 then 29 a[i,1,j]:=-maxlongint div 10 30 else 31 begin 32 a[i,1,j]:=a[i-1,1,j]+b[i,1]; 33 if a[i,1,j]<a[i-1,0,j-1]+b[i,1] then a[i,1,j]:=a[i-1,0,j-1]+b[i,1]; 34 if (j>0) and (a[i,1,j]<a[i-1,1,j-1]+b[i,1]) then a[i,1,j]:=a[i-1,1,j-1]+b[i,1]; 35 end; 36 end; 37 if a[n,0,k]>a[n,1,k] then writeln(a[n,0,k]) else writeln(a[n,1,k]); 38 halt; 39 end; 40 for i:=0 to k do for j:=0 to 4 do a[0,j,i]:=-maxlongint div 10; 41 a[0,0,0]:=0; 42 for i:=1 to n do 43 for j:=0 to k do 44 begin 45 for ii:=0 to 4 do 46 case ii of 47 0:begin 48 a[i,0,j]:=a[i-1,0,j]; 49 for jj:=1 to 4 do if a[i,0,j]<a[i-1,jj,j] then a[i,0,j]:=a[i-1,jj,j]; 50 end; 51 1..2:begin 52 a[i,ii,j]:=a[i-1,ii,j]+b[i,ii]; 53 for jj:=0 to 4 do if (j>0) and (a[i,ii,j]<a[i-1,jj,j-1]+b[i,ii]) then a[i,ii,j]:=a[i-1,jj,j-1]+b[i,ii]; 54 if a[i,ii,j]<a[i-1,3,j]+b[i,ii] then a[i,ii,j]:=a[i-1,3,j]+b[i,ii]; 55 end; 56 3:begin 57 a[i,ii,j]:=a[i-1,ii,j]+b[i,2]+b[i,1]; 58 for jj:=0 to 4 do if (j>=2) and (a[i,ii,j]<a[i-1,jj,j-2]+b[i,2]+b[i,1]) then a[i,ii,j]:=a[i-1,jj,j-2]+b[i,2]+b[i,1]; 59 for jj:=1 to 3 do if (j>=1) and (a[i,ii,j]<a[i-1,jj,j-1]+b[i,2]+b[i,1]) then a[i,ii,j]:=a[i-1,jj,j-1]+b[i,2]+b[i,1]; 60 end; 61 4:begin 62 a[i,ii,j]:=a[i-1,ii,j]+b[i,2]+b[i,1]; 63 for jj:=0 to 4 do if (j>=1) and (a[i,ii,j]<a[i-1,jj,j-1]+b[i,2]+b[i,1]) then a[i,ii,j]:=a[i-1,jj,j-1]+b[i,2]+b[i,1]; 64 end; 65 end; 66 end; 67 tot:=-maxlongint div 10; 68 for i:=0 to 4 do if tot<a[n,i,k] then tot:=a[n,i,k]; 69 writeln(tot); 70 end.
原文:http://www.cnblogs.com/HansBug/p/4296727.html