UC知道ACM求解4
来源：百度知道 编辑：UC知道 时间：2024/05/27 23:37:03
Description
Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company
is uncooperative, so he needs to pay for some of the cables required to connect his farm to
the phone system.
There are N (1 ≤ N ≤ 1,000) forlorn telephone poles conveniently numbered 1..N that are
scattered around Farmer John's property; no cables connect any them. A total of P (1 ≤ P ≤
10,000) pairs of poles can be connected by a cable; the rest are too far apart.
The i-th cable can connect the two distinct poles Ai and Bi, with length Li (1 ≤ Li ≤
1,000,000) units if used. The input data set never names any {Ai, Bi} pair more than once.
Pole 1 is already connected to the phone system, and pole N is at the farm. Poles 1 and N
need to be connected by a path of cables; the rest of the poles might be used or might not
be used.
As it turns out, the phon
Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company
is uncooperative, so he needs to pay for some of the cables required to connect his farm to
the phone system.
There are N (1 ≤ N ≤ 1,000) forlorn telephone poles conveniently numbered 1..N that are
scattered around Farmer John's property; no cables connect any them. A total of P (1 ≤ P ≤
10,000) pairs of poles can be connected by a cable; the rest are too far apart.
The i-th cable can connect the two distinct poles Ai and Bi, with length Li (1 ≤ Li ≤
1,000,000) units if used. The input data set never names any {Ai, Bi} pair more than once.
Pole 1 is already connected to the phone system, and pole N is at the farm. Poles 1 and N
need to be connected by a path of cables; the rest of the poles might be used or might not
be used.
As it turns out, the phon
这个网上都可以找到的：
核心代码如下：
int dijk()
{
memset(dps,-1,sizeof(dps));
priority_queue<elem> qu;
qu.push(elem(0,0,0));
int ax,ak,al,bx,bl,trvl;
while (!qu.empty())
{
elem tmp=qu.top();
qu.pop();
ax=tmp.py;
ak=tmp.pk;
al=tmp.pl;
if (ax==n-1) return al;
if (dps[ax][ak]<0)
{
dps[ax][ak]=al;
trvl=nodes[ax];
while (trvl>-1)
{
bx=sts[trvl].py;
bl=sts[trvl].len;
trvl=sts[trvl].next;
if (dps[bx][ak]<0) qu.push(elem(bx,ak,smax(al,bl)));
if (ak<k&&dps[bx][ak+1]<0) qu.push(elem(bx,ak+1,al));
}
}
}
return -1;
}