原题链接：Leetcode 1584. 连接所有点的最小费用
prime：
class Solution {
public:
vector<int> d;
vector<int> visit;
int cost=0;
void prime(vector<vector<int>>& points,int t,int n)
{
d[t]=0;
for(int i=0;i<n;i++)
{
int u=-1,tmp=INT_MAX;
for(int j=0;j<n;j++)
{
if(!visit[j] && d[j]<tmp)
{
u=j; tmp=d[j];
}
}
if(u==-1) return ;
visit[u]=1;
cost+=d[u];
for(int j=0;j<n;j++)
{
if(!visit[j])
{
int dis=abs(points[u][0]-points[j][0])+abs(points[u][1]-points[j][1]);
if(dis<d[j]) d[j]=dis;
}
}
}
}
int minCostConnectPoints(vector<vector<int>>& points) {
int n=points.size();
d.resize(n,INT_MAX);
visit.resize(n);
prime(points,0,n);
return cost;
}
};
kruskal： C++ STL之tuple详解
class Solution {
public:
static bool cmp(const tuple<int,int,int>& a,const tuple<int,int,int>& b)
{
return get<2>(a) < get<2>(b);
}
vector<tuple<int,int,int>> edges;
map<int,int> fa;
int findfather(int x)
{
int a=x;
while(x!=fa[x]) x=fa[x];
while(a!=fa[a])
{
int z=a;
a=fa[a];
fa[z]=x;
}
return x;
}
int cost=0,num=0;
void kruskal(int n)
{
for(auto x:edges)
{
int a=get<0>(x),b=get<1>(x),dis=get<2>(x);
int fx=findfather(a);
int fy=findfather(b);
if(fx!=fy)
{
fa[fy]=fx;
cost+=dis;
num++;
if(num==n-1) break;
}
}
}
int minCostConnectPoints(vector<vector<int>>& points) {
int n=points.size();
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
if(j>i)
{
int dis=abs(points[i][0]-points[j][0])+abs(points[i][1]-points[j][1]);
edges.push_back({
i,j,dis});
edges.push_back({
j,i,dis});
}
}
}
for(int i=0;i<n;i++) fa[i]=i;
sort(edges.begin(),edges.end(),cmp);
kruskal(n);
return cost;
}
};