NOIP2011复赛普级组 瑞士轮

瑞士轮" TITLE="NOIP2011复赛普级组 瑞士轮" />


瑞士轮" TITLE="NOIP2011复赛普级组 瑞士轮" />


瑞士轮" TITLE="NOIP2011复赛普级组 瑞士轮" />

解题思路：先读入数据，用快排对所有选手按照实力值排序，然后模拟，每打一场比赛就把赢的和输的分为两个有序数组，再进行归并，最后按要求输出即可。

程序：

const
maxn=100000;
var
a:array[0..2*maxn,1..3] of longint;
b,c:array[0..maxn,1..3] of longint;
n:longint;

procedure qsort(l,r:longint);
var
i,j,k1,k2:longint;
begin
if l>=r then
exit;
i:=l; j:=r;
k1:=a[(i+j) div
2,1];
k2:=a[(i+j) div
2,3];
repeat
while
(a[i,1]>k1)or(a[i,1]=k1)and(a[i,3]
while
(a[j,1]k2) do dec(j);
if
i<=j then

begin

a[0]:=a[i];a[i]:=a[j];a[j]:=a[0];

inc(i);dec(j);

end;
until i>j;
qsort(i,r);
qsort(l,j);
end;

procedure merge;
var
p,q,a1,i:longint;
begin
p:=1; q:=1; a1:=0;
while
(p<=n)and(q<=n) do
if
(b[p,1]>c[q,1])or(b[p,1]=c[q,1])and(b[p,3]
else
begin inc(a1); a[a1]:=c[q]; inc(q); end;
for i:=p to n do

begin

inc(a1);

a[a1]:=b[i];

end;
for i:=q to n do

begin

inc(a1);

a[a1]:=c[i];

end;
end;

procedure main;
var
r,q,i,j:longint;
begin
readln(n,r,q);
for i:=1 to n*2 do
read(a[i,1]);
readln;
for i:=1 to n*2 do
begin
read(a[i,2]);
a[i,3]:=i;
end;
qsort(1,n*2);
for i:=1 to r do
begin
for
j:=1 to n do

if a[j*2-1,2]>a[j*2,2] then begin
inc(a[j*2-1,1]); b[j]:=a[j*2-1]; c[j]:=a[j*2]; end

else begin inc(a[j*2,1]);
b[j]:=a[j*2]; c[j]:=a[j*2-1]; end;

merge;
end;
writeln(a[q,3]);
end;

begin
main;
end.

版权属于: Chris
原文地址: http://blog.sina.com.cn/s/blog_83ac6af80102v6tf.html
转载时必须以链接形式注明原始出处及本声明。