最小生成樹入門題,輸入n個座標點,輸出連接這些點的最小距離和。
解法1: Kruskal
//kruskal 1811
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define MAX 105
using namespace std;
struct Point
{
double x,y;
}point[MAX];
struct Road
{
int s,d;
double dis;
}road[MAX*MAX];
double dis(Point a,Point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool cmp(Road a,Road b)
{
return a.dis<b.dis;
}
int father[MAX];
void init()
{
for(int i=0;i<MAX;i++)
{
father[i]=i;
}
}
int find(int x)
{
return father[x]==x? x:father[x]=find(father[x]);
}
double kruskal(int index)
{
double min=0.0;
int a,b;
for(int i=0;i<index;i++)
{
a=find(road[i].s);
b=find(road[i].d);
if(a!=b)
{
father[a]=b;
min+=road[i].dis;
}
}
return min;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int i,j;
memset(point,0,sizeof(point));
for(i=0;i<n;i++)
{
scanf("%lf %lf",&point[i].x,&point[i].y);
}
int index=0;
for(i=0;i<n;i++)
{
for(j=i+1;j<=n-1;j++)
{
road[index].s=i;
road[index].d=j;
road[index++].dis=dis(point[i],point[j]);
}
}
sort(road,road+index,cmp);
init();
printf("%.2lf\n",kruskal(index));
}
return 0;
}
解法2: Prim
//1811 Prim
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define INF 0xffffff
using namespace std;
struct pos
{
double x,y;
}f[105];
double map[105][105];
bool visited[105];
double dist[105];
double dis(pos a,pos b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main()
{
int n;
while(scanf(" %d",&n)!=EOF)
{
int i,j;
for(i=0;i<n;i++)
{
scanf("%lf %lf",&f[i].x,&f[i].y);
}
memset(map,0,sizeof(map));
memset(visited,0,sizeof(visited));
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
map[i][j]=dis(f[i],f[j]);
}
}
//memset(dist,100000,sizeof(dist));
//memset是按字節賦值,不要亂用。。。
for(int i=0;i<105;i++)
{
dist[i]=INF;
}
double ans=0;
dist[0]=0;
for(int i=0;i<n;i++)
{
int k=0;
double min=INF;
//找到一個能使和最小的點 將該點加入集合
for(int j=0;j<n;j++)
{
if(!visited[j] && dist[j]<min)
{
min=dist[j];
k=j;
}
}
ans+=min;
visited[k]=1;
//對錶進行更新
for(int j=0;j<n;j++)
{
if(dist[j]>map[k][j])
{
dist[j]=map[k][j];
}
}
}
printf("%.2lf\n",ans);
}
return 0;
}
解法3: Dijkstra
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define INF 0xffffff
using namespace std;
struct pos
{
double x,y;
}f[105];
double map[105][105];
bool visited[105];
double dis(pos a,pos b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int i,j;
memset(f,0,sizeof(f));
for(i=0;i<n;i++)
{
scanf("%lf %lf",&f[i].x,&f[i].y);
}
memset(map,INF,sizeof(map));
memset(visited,0,sizeof(visited));
for(i=0;i<n-1;i++)
{
for(j=i+1;j<n;j++)
{
map[i][j]=map[j][i]=dis(f[i],f[j]);
}
}
double ans=0;
int temp;
//0爲起點 找點
visited[0]=1;
for(int k=0;k<n-1;k++)
{
double s=INF;
//找到一個能使和最小的點 將該點加入集合
for(int i=1;i<n;i++)
{
if(!visited[i] && map[0][i]<s)
{
s=map[0][i];
temp=i;
}
}
ans+=s;
visited[temp]=1;
//對錶進行更新
for(int j=1;j<n;j++)
{
if(!visited[j] && map[temp][j]<map[0][j])
{
map[0][j]=map[temp][j];
}
}
}
printf("%.2lf\n",ans);
}
return 0;
}