[NDK 安置诸侯]

[题目来源]：2011/11/04模拟赛

[关键字]：递推

[题目大意]：在一个类似菱形的地图里放置诸侯，保证每一行每一列都只有一个诸侯。

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[分析]：先把地图转化成，再用递推。可以很容易的求出方程：f[i,j]=∑f[k,j-1]*(l[i]-(j-1))其中f[i,j]为前i列放j个，l[i]是第i列有多少个。

[代码]：

View Code

 1 {$i-,s-,q-,r-}
 2 var
 3   ans, k, n, i, j, m, temp: longint;
 4   f: array[0..200,0..200] of longint;
 5 
 6 function min(x, y: longint):longint;
 7 begin
 8   if x < y then exit(x) else exit(y);
 9 end;
10 
11 procedure print(x: longint);
12 begin
13   writeln(x);
14   halt;
15 end;
16 
17 begin
18   readln(n,m);
19   if m = 0 then print(1);
20   if m >= 2*n-1 then print(0);
21   fillchar(f,sizeof(f),0);
22   f[0,0] := 1;
23   for i := 1 to n do
24     if odd(i) then f[i,1] := i else f[i,1] := i-1;
25   for i := 1 to 2*n-1 do
26     begin
27       if odd(i) then temp := i else temp := i-1;
28       for j := 2 to min(i,m) do
29         for k := j-1 to i-1 do
30           f[i,j] := (f[i,j]+f[k,j-1]*(temp-j+1)) mod 504;
31     end;
32   for i := m to 2*n-1 do ans := (ans+f[i,m]) mod 504;
33   print(ans);
34 end.