POJ1287
输入有重边,Kruskal完成。实际上,数据很小0.0
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define Maxm 100000
#define Maxn 1000
using namespace std;
struct Edge{
int u,v,len;
}edge[Maxm];
int fa[Maxn];
int n,m;
bool cmp(Edge a,Edge b){
return a.len<b.len;
}
int getfather(int x){
if (x!=fa[x]) return fa[x]=getfather(fa[x]);
return fa[x];
}
int Kruskal(){
int ans=0,u,v;
for (int i=1;i<=n;i++) fa[i]=i;
for (int i=1;i<=m;i++){
u=getfather(edge[i].u);v=getfather(edge[i].v);
if (u==v) continue;
fa[v]=u;
ans+=edge[i].len;
}
return ans;
}
int main(){
while (scanf("%d",&n),n!=0){
scanf("%d",&m);
for (int i=1;i<=m;i++){
scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].len);
}
sort(edge+1,edge+1+m,cmp);
printf("%d\n",Kruskal());
}
}
POJ1861
求最小生成树。输出路径
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
struct Edge{
int u,v,len;
}edge[20000];
int top,n,m,ans;
int f[1500],s[20000];
bool cmp(Edge x,Edge y)
{
return x.len<y.len;
}
int getfather(int x)
{
if (x!=f[x]) f[x]=getfather(f[x]);
return f[x];
}
int kruscal()
{
int u,v;
for (int i=0;i<=n;i++) f[i]=i;
top=0;
for (int i=1;i<=m;i++)
{
u=getfather(edge[i].u);v=getfather(edge[i].v);
if (u==v) continue;
f[v]=u;
ans=edge[i].len;
top++;s[top]=i;
if (top>=n-1) break;
}
return 0;
}
int main()
{
while (scanf("%d%d",&n,&m)!=EOF)
{
for (int i=1;i<=m;i++)
scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].len);
sort(edge+1,edge+m+1,cmp);
kruscal();
printf("%d\n",ans);
printf("%d\n",top);
for (int i=1;i<=top;i++)
printf("%d %d\n",edge[s[i]].u,edge[s[i]].v);
}
}
POJ2349
题目大意:求允许S个连通块的最小生成树。
分析:不要执着于S个连通块,换个角度理解就是S-1条边免费。于是,我们求出最小生成树后,输出第S+1大的边的权值就能得到答案了。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#define Maxn 501
#define Maxm 250050
using namespace std;
struct Node{
double x,y;
}b[Maxn];
struct Edge{
int u,v;
double len;
}edge[Maxm];
int fa[Maxn];
int top,n,s;
bool cmp(Edge a,Edge b){
return a.len<b.len;
}
double Distance(Node a,Node b){
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int getfather(int x){
if (x!=fa[x]) return fa[x]=getfather(fa[x]);
return fa[x];
}
double Kruskal(){
int u,v;
double ans=0;
for (int i=1;i<=n;i++) fa[i]=i;
int cnt=0;
for (int i=1;i<=top-s;i++){
u=getfather(edge[i].u);v=getfather(edge[i].v);
if (u==v) continue;
fa[v]=u;cnt++;
if (cnt==n-1-s){
ans=edge[i].len;
break;
}
}
return ans;
}
int main(){
int cases;
scanf("%d",&cases);
for (int cas=1;cas<=cases;cas++){
scanf("%d%d",&s,&n);s--;
for (int i=1;i<=n;i++){
scanf("%lf%lf",&b[i].x,&b[i].y);
}
top=0;
for (int i=1;i<=n;i++){
for (int j=i+1;j<=n;j++){
top++;edge[top].u=i;edge[top].v=j;
edge[top].len=Distance(b[i],b[j]);
}
}
sort(edge+1,edge+1+top,cmp);
printf("%.2f\n",Kruskal());
}
return 0;
}