uva 10034 Freckles (kruskal||prim)

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题目上只给的坐标，没有给出来边的长度，无论是prim算法还是kruskal算法我们都需要知道边的长度来操作。

这道题是浮点数，也没啥大的区别，处理一下就可以了。有关这两个算法的介绍前面我已经写过了，就不在多写了

prim算法：

<span style="font-family:Courier New;font-size:18px;">#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<vector>#include<set>#include<string>#include<algorithm>#include<climits>using namespace std;struct node{double i,j;}g[105];double gra[105][105];double dist(double a,double b,double c,double d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}int n,cnt=0,T;void prim(){int visit[105],now,i,j;double dis[105];double Min;memset(visit,0,sizeof(visit));for(i=1; i<=n; i++)dis[i] = INT_MAX;visit[1] = 1, dis[1] = 0, now = 1;//now都是当前新加的点 for(i=1; i<=n; i++){for(j=1; j<=n; j++){if(!visit[j] && dis[j]>gra[now][j])//用新加的点来更新其他点到此集合的距离 dis[j] = gra[now][j];}Min = INT_MAX;for(j=1; j<=n; j++){if(!visit[j] && dis[j] < Min)//每次都找到距离最小的点，加进去 Min = dis[now = j];}visit[now] = 1;} double sum = 0;for(i=1; i<=n; i++){sum += dis[i];}printf("%.2lf\n",sum);if(cnt!= T)//注意每两个输出案例之间都有一个换行 cout << endl;}int main(){int i,j;cin >> T;while(cin >> n){cnt ++;for(i=1; i<=n; i++){cin >> g[i].i >> g[i].j;} memset(gra,0,sizeof(gra));for(i=1; i<=n; i++){for(j=i+1; j<=n; j++){gra[i][j] = gra[j][i] = dist(g[i].i,g[i].j,g[j].i,g[j].j);}}prim();}return 0;}</span>

kruskal算法:

<span style="font-family:Courier New;font-size:18px;">#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<vector>#include<set>#include<string>#include<algorithm>#include<climits>using namespace std;struct node{int i,j;double len;}gra[10005];struct node1{double i,j;}g[105];int p[105];double dist(double a,double b,double c,double d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}int cmp(const void *a,const void *b){return (((node *)a)->len - ((node *)b)->len > 0) ? 1:-1;}int n,cnt=0,T,k;int find(int x){return x == p[x]? x: p[x] = find(p[x]);}void kruskal(){double sum = 0;int i;for(i=1; i<k; i++){int x = find(gra[i].i);int y = find(gra[i].j);if(x!=y){sum += gra[i].len;p[x] = y;}}printf("%.2f\n",sum);if(cnt != T)cout << endl;}int main(){int i,j;cin >> T;while(cin >> n){cnt ++;for(i=1; i<=n; i++){cin >> g[i].i >> g[i].j;} memset(gra,0,sizeof(gra));k=1;for(i=1; i<=n; i++){for(j=i+1; j<=n; j++){gra[k].len = dist(g[i].i,g[i].j,g[j].i,g[j].j);gra[k].i = i;gra[k].j = j;k++;}}//prim();for(i=1; i<=n; i++)p[i] = i;qsort(gra+1,k-1,sizeof(gra[0]),cmp);kruskal();}return 0;}</span>

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