[BZOJ1084] [SCOI2005]最大子矩阵

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传送门

http://www.lydsy.com/JudgeOnline/problem.php?id=1084

题目大意

给定一个矩阵选取其中不相交的t个矩阵使其权值和最大

题解

m=1或2
m=1时
dp[i,j]=max{dp[i−1,j],max{dp[k,j−1]+sum[i]−sum[k]}}  (0<=k<i)
m=2时
dp[i,j,k]=max{dp[i−1,j,k],dp[i,j−1,k]}
dp[i,j,k]=max{dp[i,j,k],dp[l,j,k−1]+sum1[i]−sum1[l]}  (0<=l<i)
dp[i,j,k]=max{dp[i,j,k],dp[i,l,k−1]+sum2[j]−sum2[l]}  (0<=l<j)
dp[i,j,k]=max{dp[i,j,k],dp[l,l,k−1]+sum1[i]−sum1[l]+sum2[i]−sum2[l]}  (i=j且0<=l<i)

var sum1,sum2:array[0..200]of longint; dp1:array[0..200,0..200]of longint; dp2:array[0..200,0..200,0..200]of longint; i,j,k,l:longint; n,m,t,a,b:longint;function max(a,b:longint):longint;begin if a>b then exit(a) else exit(b);end;begin readln(n,m,t); if m=1 then begin  sum1[0]:=0;  for i:=1 to n do   begin    readln(a);    sum1[i]:=sum1[i-1]+a;   end;  for i:=1 to n do   for j:=1 to t do    begin     dp1[i,j]:=dp1[i-1,j];     for k:=0 to i-1 do      dp1[i,j]:=max(dp1[i,j],dp1[k,j-1]+sum1[i]-sum1[k]);    end;  writeln(dp1[n,t]); end else begin  sum1[0]:=0; sum2[0]:=0;  for i:=1 to n do   begin    readln(a,b);    sum1[i]:=sum1[i-1]+a; sum2[i]:=sum2[i-1]+b;   end;  for i:=1 to n do   for j:=1 to n do    for k:=1 to t do     begin      dp2[i,j,k]:=max(dp2[i-1,j,k],dp2[i,j-1,k]);      for l:=0 to i-1 do       dp2[i,j,k]:=max(dp2[i,j,k],dp2[l,j,k-1]+sum1[i]-sum1[l]);      for l:=0 to j-1 do       dp2[i,j,k]:=max(dp2[i,j,k],dp2[i,l,k-1]+sum2[j]-sum2[l]);      if i=j then      for l:=0 to i-1 do       dp2[i,j,k]:=max(dp2[i,j,k],dp2[l,l,k-1]+sum1[i]-sum1[l]+sum2[i]-sum2[l]);     end;  writeln(dp2[n,n,t]); end;end.

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