POJ-3662(Telephone Lines)
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 phone company is willing to provide Farmer John with K (0 ≤ K < N) lengths of cable for free. Beyond that he will have to pay a price equal to the length of the longest remaining cable he requires (each pair of poles is connected with a separate cable), or 0 if he does not need any additional cables.
Determine the minimum amount that Farmer John must pay.
Input
* Line 1: Three space-separated integers: N, P, and K
* Lines 2..P+1: Line i+1 contains the three space-separated integers: Ai, Bi, and Li
Output
* Line 1: A single integer, the minimum amount Farmer John can pay. If it is impossible to connect the farm to the phone company, print -1.
Sample Input
5 7 1 1 2 5 3 1 4 2 4 8 3 2 3 5 2 9 3 4 7 4 5 6
Sample Output
4
我的答案：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct edge
{
int a,b,l;
}bian[10100];
int map[1010][1010];
int vis[1010];
int dist[1010];
int inf=0x7ffffff;
bool cmp(struct edge a,struct edge b)
{
return a.l<b.l;
}
void build(int l,int n,int p)
{
int i,j;
for(i=0;i<=n;++i)
{
for(j=0;j<=n;++j)
{
map[i][j]=inf;
}
}
for(i=0;i<p;++i)
{
int a,b;
a=bian[i].a;
b=bian[i].b;
if(bian[i].l<=l)
{
map[a][b]=map[b][a]=0;
}else map[a][b]=map[b][a]=1;
}
}
int xun(int n)//在剩余的节点中寻找到1最近的
{
int xia=-1,i;
for(i=1;i<=n;i++)
if(vis[i]==0&&(xia==-1||dist[i]<dist[xia]))
xia=i;
return xia;
}
int djk(int n)
{
int i,j,just;
for(i=1;i<=n;i++)//初始化
dist[i]=map[1][i];
memset(vis,0,sizeof(vis));
vis[1]=1;
for(i=2;i<=n;i++)
{
just=xun(n);
vis[just]=1;//just最近，将他标为已找到最短路径
for(j=1;j<=n;j++)//更新剩余的
if(vis[j]==0&&dist[j]>dist[just]+map[just][j])
dist[j]=dist[just]+map[just][j];
}
return dist[n];
}
int bin(int n,int p,int k)
{
int head,tail;
head=0;
tail=p-1;
int mid;
while(head<=tail)
{
int mid=(head+tail)>>1;
build(bian[mid].l,n,p);
if(djk(n)<=k)//求出最小的
{
tail=mid-1;
}
else head=mid+1;
}
return head;
}
int main()
{
int i,n,p,k;
while(scanf("%d%d%d",&n,&p,&k)!=EOF)
{
for(i=0;i<p;i++)
{
scanf("%d%d%d",&bian[i].a,&bian[i].b,&bian[i].l);
}
sort(bian,bian+p,cmp);
build(0,n,p);
int duan=djk(n);
if(duan==inf)
{
printf("-1\n");
}else
{
if(duan<=k)
printf("0\n");
else printf("%d\n",bian[bin(n,p,k)].l);
}
}
return 0;
}