topic
| Time Limit: 1000MS | Memory Limit: 65536K |
|---|---|
| Total Submissions: 9674 | Accepted: 3483 |
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
Source
USACO 2008 January Silver
code
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
template<typename T>inline void read(T &x)
{
x=0;
T f=1,ch=getchar();
while (!isdigit(ch) && ch^'-') ch=getchar();
if (ch=='-') f=-1, ch=getchar();
while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48), ch=getchar();
x*=f;
}
int ver[maxn],edge[maxn],Next[maxn],len[maxn],head[maxn],tot;
inline void add(int x,int y,int z)
{
ver[++tot]=y,len[tot]=z,Next[tot]=head[x],head[x]=tot;
}
int n,p,k;
int dist[maxn],vis[maxn];
inline int spfa(int s)
{
memset(dist,0x3f,sizeof(dist));
memset(vis,0,sizeof(vis));
queue<int>q;
dist[s]=0,vis[s]=1;
q.push(s);
while (!q.empty())
{
int x=q.front();
q.pop();
vis[x]=0;
for (int i=head[x];i;i=Next[i])
{
int y=ver[i],z=edge[i];
if (dist[y]>dist[x]+z)
{
dist[y]=dist[x]+z;
if (!vis[y]) q.push(y),vis[y]=1;
}
}
}
return dist[n]<=k;
}
inline int check(int mid)
{
for (int x=1;x<=n;++x)
for (int i=head[x];i;i=Next[i])
if (len[i]<=mid) edge[i]=0;
else edge[i]=1;
return spfa(1);
}
int main()
{
int r=-1,ans=-1;
read(n);read(p);read(k);
for (int i=1;i<=p;++i)
{
int x,y,z;
read(x);read(y);read(z);
add(x,y,z);add(y,x,z);
r=max(r,z);
}
int l=0;
while (l<=r)
{
int mid=(l+r)>>1;
if (check(mid)) r=mid-1,ans=mid;
else l=mid+1;
}
printf("%d\n",ans);
return 0;
}
所谓的双端队列优化，吸了O2才过，也许是我代码能力太菜了。
评测结果
// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
template<typename T>inline void read(T &x)
{
x=0;
T f=1,ch=getchar();
while (!isdigit(ch) && ch^'-') ch=getchar();
if (ch=='-') f=-1, ch=getchar();
while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48), ch=getchar();
x*=f;
}
int ver[maxn],edge[maxn],Next[maxn],head[maxn],tot;
inline void add(int x,int y,int z)
{
ver[++tot]=y,edge[tot]=z,Next[tot]=head[x],head[x]=tot;
}
int n,p,k,maxmum;
int dist[maxn],vis[maxn];
inline int bfs(int mid)//bfs求最短路
{
memset(dist,0,sizeof(dist));
memset(vis,0,sizeof(vis));
deque<int>q;
dist[1]=0,vis[1]=1;
q.push_back(1);
while (!q.empty())
{
int x=q.front();
q.pop_front();
for (int i=head[x];i;i=Next[i])
{
int y=ver[i];
if (!vis[y] || dist[y]>=dist[x]+1)
{
if (edge[i]<=mid)//将小于等于mid的边看做权值为零的边
{
vis[y]=1,q.push_front(y);
dist[y]=dist[x];
}
else//然后将大于mid的边看做权值为1的边
{
vis[y]=1,q.push_front(y);
dist[y]=dist[x]+1;
}
}
}
}
if (dist[n]>k) return 0;//若最短路的长度大于k，则要连接的对数大于k对，在[mid+1,r]中继续二份查找
else return 1;//若最短路的长度小于k，则要连接的对数比k小，在[l,mid]中继续二份查找
}
int main()
{
read(n);read(p);read(k);
while (p--)
{
int x,y,z;
read(x);read(y);read(z);
add(x,y,z);add(y,x,z);
maxmum=max(maxmum,z);
}
int l=1,r=maxmum;
while (l<r)//可以二分最大的花费mid，mid属于[l,r](l为最小花费，r为最大花费)
{
int mid=(l+r)>>1;
if (bfs(mid)) r=mid;
else l=mid+1;
}
if (l!=1) printf("%d\n",l);//最终，l即为所求
else puts("-1");
return 0;
}