摘要:
1 program Xiong; 2 var 3 count,father,data:array[0..10000]of longint; 4 son:array[0..10000,0..1]of longint; 5 root,n,m,t,i:longint; 6 7 procedure rotate(x,how:longint); 8 var 9 up,uup:longint; 10 begin 11 up:=father[x]; 12 uup:=father[up]; 1... 阅读全文
posted @ 2012-04-29 18:35
不二的笨笨
阅读(177)
评论(0)
推荐(0)

浙公网安备 33010602011771号