Time Limit: 1000MS | Memory Limit: 30000K | |||
Total Submissions: 17321 | Accepted: 6970 | Special Judge |
Description
Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem — not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections.
You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied.
Input
Output
Sample Input
4 6
1 2 1
1 3 1
1 4 2
2 3 1
3 4 1
2 4 1
Sample Output
1
4
1 2
1 3
2 3
3 4
Source
題意:
求最小生成樹的最大權值,並輸出經過的邊數和路徑
代碼:
(prime算法)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int dis[1005][1005],dist[1005],vis[1005];
int n,m,maxk;
int f[1005];
void prime(){
int sum=0;
maxk=0;
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++){
dist[i]=dis[1][i];
f[i]=1;//記錄父親節點
}
vis[1]=1;
int m=n-1;
while(m--){
int minx=INF;
int p=0;
for(int i=1;i<=n;i++){
if(!vis[i]&&dist[i]<minx){
minx=dist[i];
p=i;
}
}
vis[p]=1;
if(p==0)break;
sum+=minx;
maxk=max(maxk,minx);
for(int i=1;i<=n;i++){
if(!vis[i]&&dist[i]>dis[p][i]){
f[i]=p; //更新父親節點
dist[i]=dis[p][i];
}
}
}
}
int main(){
scanf("%d%d",&n,&m);
memset(dis,INF,sizeof(dis));
for(int i=1;i<=m;i++){
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
dis[a][b]=dis[b][a]=min(dis[a][b],c);
}
prime();
printf("%d\n%d\n",maxk,n-1);
for(int i=2;i<=n;i++)printf("%d %d\n",f[i],i);
return 0;
}
(kruscal算法)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int n,m;
int maxk,cnt;
int ax[15000],ay[15000],f[15000];
struct data{
int x,y,c;
}edge[15000];
bool cmp(data a,data b){
return a.c<b.c;
}
void pre(){
for(int i=0;i<=n;i++)f[i]=i;
}
int find(int x){
return f[x]==x?x:find(f[x]);
}
void kruscal(){
int sum=0;
cnt=0;
sort(edge,edge+m,cmp);
for(int i=0;i<m;i++){
int x=find(edge[i].x);
int y=find(edge[i].y);
if(x!=y){
maxk=max(maxk,edge[i].c);
sum++;
f[x]=y;
ax[cnt]=edge[i].x;
ay[cnt++]=edge[i].y;
}
if(sum>=n-1)break;
}
}
int main(){
scanf("%d%d",&n,&m);
int a,b,c;
for(int i=0;i<m;i++){
scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].c);
}
maxk=0;
pre();
kruscal();
printf("%d\n%d\n",maxk,cnt);
for(int i=0;i<cnt;i++){
printf("%d %d\n",ax[i],ay[i]);
}
return 0;
}