Noip2003提高组解题报告1.docVIP

Noip 2003 提高组 解题报告题一??? 神经网络 分析：??? 这一题有很多种方法，数据也较弱。我考虑到其中的层次关系，用类似于广搜的方法，但是注意到每个结点不仅仅是由当前head节点扩展，而是当前队列全部节点。 源程序：program network1;const maxn=100;var c,u:array[1..maxn]of longint; //c，状态数组；u，阈值数组??? w:array[1..maxn,1..maxn]of longint; //表示是否连通(有向图）??? anstp,head,tail,n,i,j,p,k,x,y:longint;??? q:array[1..10000]of record dep,num:longint; end; //队列记录两个变量（层次与号码）??? ct:array[1..maxn]of boolean; //判重数组??? ans:array[1..maxn]of longint; //答案，记得排序，故先用数组暂存??? tag:boolean; //判断是否输出NULL procedure qsort(s,t:longint); //快速排序var i,j,d,x:longint;begin????? i:=s; j:=t; x:=ans[(s+t)shr 1];????? repeat?????? while ans[i]x do inc(i);?????? while ans[j]x do dec(j);?????? if i=j??????? then begin d:=ans[i]; ans[i]:=ans[j]; ans[j]:=d; inc(i); dec(j); end;????? until ij;????? if sj then qsort(s,j); if it then qsort(i,t);end; begin????? assign(input,network.in); reset(input); assign(output,network.out); rewrite(output); ?????? {init}?????? readln(n,p);?????? for i:=1 to n do readln(c[i],u[i]);?????? fillchar(w,sizeof(w),0); fillchar(ct,sizeof(ct),false);?????? for i:=1 to p do begin readln(x,y,k); w[x,y]:=k; w[y,x]:=k; end;???????????? {main}?????? head:=0; tail:=0;?????? for i:=1 to n do??????? if c[i]0// 对首层的操作，可以传递兴奋则进队???????? then????????? begin?????????? ct[i]:=true; inc(tail); q[tail].dep:=1; q[tail].num:=i;????????? end;?????? for i:=1 to n do if c[i]=0 then c[i]:=-u[i]; //先确定起始值?????? repeat??????? x:=head;??????? repeat???????? inc(x);???????? if c[q[x].num]=0 then continue; //若兴奋性=0，则不会向下传递，枚举下一个???????? for i:=1 to n do????????? if not ct[i] and (w[q[x].num,i]0) //没有入队，存在有向边连接?????????? then c[i]:=c[i]+c[q[x].num]*w[q[x].num,i];??????? until x=tail;?? //下一层不仅仅取决于上一层的某一个元素，它来源于上一层的全部元素。??????? x:=tail;? //恢复尾指针??????? repeat???????? inc(head);???????? for i:=1 to n do????????? if not ct[i] and (w[i,q[head].num]0)?????????? then begin inc(tail); ct[i]:=true; q[tail].num:=i; q[tail].dep:=q[head].dep+1; end;?????????? //拓展队列，注意无论是