[BZOJ1087][SCOI2005]互不侵犯King解题报告|状压DP

在N×N的棋盘里面放K个国王，使他们互不攻击，共有多少种摆放方案。国王能攻击到它上下左右，以及左上左下右上右下八个方向上附近的各一个格子，共8个格子。

 

　　好像若干月前非常Naive地去写过DFS...

　　然后其实作为状压DP是一道非常好的题啦><

　　感觉直接无脑搞时间是下不来的 做了好几道预处理

　　使得最后DP的过程中没有任何一条转移是无用的

　　

program bzoj1087;
var i,x,n,k,j,p,q,t1,t2:longint;
    ans:int64;
    a:array[-1..512]of longint;
    b:array[-1..9,-1..100]of longint;
    c:array[-1..512,-1..512]of longint;
    f:array[0..1,-1..512,-1..512]of int64;

function check(x:longint):boolean;
var las:longint;
begin
    las:=0;
    while x<>0 do
    begin
        if (x and 1=1)and(las=1) then exit(false);
        las:=x and 1;
        x:=x >> 1;
    end;
    exit(true);
end;

function count(x:longint):longint;
var tem:longint;
begin
    tem:=0;
    while x<>0 do
    begin
        inc(tem,x and 1);
        x:=x >> 1;
    end;
    exit(tem);
end;

function ok(x,y:longint):boolean;
var i:longint;
    tmp:array[0..1,-1..10]of longint;
begin
    for i:=8 downto 0 do tmp[0,i]:=x >> i and 1;
    for i:=8 downto 0 do tmp[1,i]:=y >> i and 1;
    for i:=0 to 8 do if (tmp[0,i]+tmp[1,i]=2)or(tmp[0,i]+tmp[1,i-1]=2)or(tmp[0,i]+tmp[1,i+1]=2) then exit(false);
    exit(true);
end;

begin
    fillchar(b,sizeof(b),0);
    fillchar(a,sizeof(a),0);
    fillchar(c,sizeof(c),0);
        readln(n,k);
        for i:=0 to 1 << n-1 do if check(i) then
    begin
            x:=count(i);
        inc(b[x,0]);b[x,b[x,0]]:=i;
        inc(a[0]);a[a[0]]:=i;
    end;
    for i:=1 to a[0] do
        for j:=1 to a[0] do if ok(a[i],a[j]) then begin inc(c[a[i],0]);c[a[i],c[a[i],0]]:=a[j];end;
    fillchar(f,sizeof(f),0);
    for i:=0 to k do
        for j:=1 to b[i,0] do f[1,i,b[i,j]]:=1;
    for i:=2 to n do
    begin
        f[0]:=f[1];
        fillchar(f[1],sizeof(f[1]),0);
        for j:=0 to k do
            for p:=0 to (n+1) >> 1 do if p<=j then
            begin
                q:=j-p;
                for t1:=1 to b[p,0] do
                    for t2:=1 to c[b[p,t1],0] do inc(f[1,j,b[p,t1]],f[0,q,c[b[p,t1],t2]]);
            end;
    end;
    ans:=0;
    for i:=1 to a[0] do inc(ans,f[1,k,a[i]]);
    writeln(ans);
end.