Three Lines (USACO 2012 US Open, Bronze Division Problem 2)

原题

{
http://usaco.org/index.php?page=viewproblem2&cpid=131
USACO 2012 US Open, Bronze Division
Problem 2. Three Lines 

by hcy 2012.10.23
}
const m=50000;
type arr=array[1..5,1..2] of longint;
var n,i,j,max:longint;
a:array[0..m,1..2] of longint;
b:arr;
function check(line,k:longint;c:arr):boolean;
var i:integer;
begin
    for i:=1 to line-1 do
        if ((c[i,1]=1) and (c[i,2]=a[k,1])) or ((c[i,1]=2) and (c[i,2]=a[k,2])) then exit(true);
    exit(false);    
end;
function getk(line,k:longint;c:arr):longint;
var i,j:longint;
f:boolean;
begin
    for j:=k to n do
    begin
        f:=false;
        for i:=1 to line-1 do
            if ((c[i,1]=1) and (c[i,2]=a[j,1])) or ((c[i,1]=2) and (c[i,2]=a[j,2])) then begin f:=true;break;end;
        if not f then exit(j);    
    end;
    exit(n+1);//to the end
end;
procedure success;
var i:longint;
begin
    writeln(1);
    close(input);close(output);
    halt;
end;
procedure find(line,k:longint;c:arr);
var i,s,t,nextk:longint;
tc:arr;
begin    
    if (k>n) then
        if (line<=4) then success
        else exit;
    if line>4 then exit;
    if check(line,k,c)    then 
    begin
        nextk:=getk(line,k+1,c);// this is important.if u use find(line,k+1,c) then it may occur 'stack overflow error' for recursion too many times
        find(line,nextk,c)
    end
    else
    begin
        //x
        tc:=c;
        tc[line,1]:=1;
        tc[line,2]:=a[k,1];
        find(line+1,k+1,tc);    
        //y    
        tc:=c;
        tc[line,1]:=2;
        tc[line,2]:=a[k,2];
        find(line+1,k+1,tc);
    end;
end;
begin
    assign(input,'security.in');reset(input);
    assign(output,'security.out');rewrite(output);
    readln(n);
    for i:=1 to n do readln(a[i,1],a[i,2]);
    find(1,1,b);
    writeln(0);
    close(input);close(output);
end.

 or:

{
http://usaco.org/index.php?page=viewproblem2&cpid=131
USACO 2012 US Open, Bronze Division
Problem 2. Three Lines 

by hcy 2012.10.23
}
const m=50000;
type arr=array[1..5,1..2] of longint;
var n,i,j,max:longint;
a:array[0..m,1..2] of longint;
b:arr;
function getk(line,k:longint;c:arr):longint;//find next position k which is not on the lines
var i,j:longint;
f:boolean;
begin
    for j:=k to n do
    begin
        f:=false;
        for i:=1 to line-1 do
            if ((c[i,1]=1) and (c[i,2]=a[j,1])) or ((c[i,1]=2) and (c[i,2]=a[j,2])) then begin f:=true;break;end;
        if not f then exit(j);    
    end;
    exit(n+1);//to the end
end;
procedure success;
var i:longint;
begin
    writeln(1);
    close(input);close(output);
    halt;
end;
procedure find(line,k:longint;c:arr);
var i,s,t,nextk:longint;
tc:arr;
begin    
    if (k>n) then
        if (line<=4) then success
        else exit;
    if line>4 then exit;
    begin        
        //x
        tc:=c;
        tc[line,1]:=1;
        tc[line,2]:=a[k,1];
        nextk:=getk(line+1,k+1,tc);// this is important.if u use find(line,k+1,c) then it may occur 'stack overflow error' for recursion too many times
        find(line+1,nextk,tc);    
        //y    
        tc:=c;
        tc[line,1]:=2;
        tc[line,2]:=a[k,2];
        nextk:=getk(line+1,k+1,tc);
        find(line+1,nextk,tc);
    end;
end;
begin
    assign(input,'security.in');reset(input);
    assign(output,'security.out');rewrite(output);
    readln(n);
    for i:=1 to n do readln(a[i,1],a[i,2]);
    find(1,1,b);
    writeln(0);
    close(input);close(output);
end.

 tip:

递归写起来很简单，但一定记得尽量减少递归次数！否则堆栈溢出！