[POJ 2411] Mondriaan's Dream

最新推荐文章于 2022-07-18 10:07:01 发布

原创最新推荐文章于 2022-07-18 10:07:01 发布 · 573 阅读

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动态规划—状压DP 专栏收录该内容

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本文介绍了一道经典的编程竞赛题目——使用1*2骨牌覆盖n*m大小的棋盘，求所有可能的覆盖方式数量。通过动态规划的方法，利用检查函数验证骨牌放置的有效性，并通过递推公式计算总数。

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传送门

http://poj.org/problem?id=2411

题目大意

1*2骨牌,覆盖n*m棋盘的方式总数

题解

http://blog.youkuaiyun.com/xiaozhuaixifu/article/details/10221341这篇比较清楚

var
 dp:array[0..200,0..3000]of int64;
 i,j,k:longint;
 n,m:longint;
function check1(a:longint):longint;
var i:longint;
begin
 i:=1;
 while i<=m do
  begin
   if a and (1<<(i-1))=0
   then inc(i)
   else
    if (a and (1<<i)<>0)and(i<=m-1)
    then inc(i,2)
    else exit(0);
  end;
 exit(1);
end;

function check(a,b:longint):longint;
var i,j,k:longint;
begin
 i:=1;
 while i<=m do
  begin
   if a and (1<<(i-1))=0
   then begin
    if b and (1<<(i-1))=0
    then exit(0)
    else inc(i);
   end
   else begin
    if b and (1<<(i-1))=0
    then inc(i)
    else
     if (i<=m-1)and(a and (1<<i)<>0)and(b and (1<<i)<>0)
     then inc(i,2)
     else exit(0);
   end;
  end;
 exit(1);
end;

begin
 while not eof do begin
 readln(n,m); if (n=0)and(m=0) then break;
 if n<m then begin k:=n; n:=m; m:=k; end;
 fillchar(dp,sizeof(dp),0);
 for j:=0 to (1<<m)-1 do
  if check1(j)=1
  then dp[1,j]:=1;
 for i:=2 to n do
  for j:=0 to (1<<m)-1 do
   for k:=0 to (1<<m)-1 do
    if check(j,k)=1
    then dp[i,j]:=dp[i,j]+dp[i-1,k];
 writeln(dp[n,(1<<m)-1]); end;
end.