UVa 10034 - Freckles (最小生成树模板题)

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链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=975

题目：

Problem A: Freckles

In an episode of the Dick Van Dyke show, little Richie connects the freckles on his Dad's back to form a picture of the Liberty Bell. Alas, one of the freckles turns out to be a scar, so his Ripley's engagement falls through.

Consider Dick's back to be a plane with freckles at various (x,y) locations. Your job is to tell Richie how to connect the dots so as to minimize the amount of ink used. Richie connects the dots by drawing straight lines between pairs, possibly lifting the pen between lines. When Richie is done there must be a sequence of connected lines from any freckle to any other freckle.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The first line contains 0 < n <= 100, the number of freckles on Dick's back. For each freckle, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the freckle.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the freckles.

Sample Input

13
1.0 1.0
2.0 2.0
2.0 4.0

Sample Output

3.41

分析与总结：

赤裸裸的最小生成树，模板题

代码：

1. Kruskal

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define N 105
double coord[N][2], w[N][N];
int n, pos;
int f[N*N], rank[N*N];struct Edge{int u, v;double val;friend bool operator<(const Edge&a,const Edge&b){return a.val < b.val;}
}arr[N*N];inline double getDist(double x1,double y1,double x2,double y2){return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); 
}void init(){for(int i=0; i<N*N; ++i)f[i]=i, rank[i]=0;
}
int find(int x){int i,j=x;while(j!=f[j]) j=f[j];while(x!=j){i=f[x]; f[x]=j; x=i;}return j;
}
bool Union(int x,int y){int a=find(x),b=find(y);if(a==b)return false;if(rank[a]>rank[b])f[b]=a;else{if(rank[a]==rank[b])++rank[b];f[a] = b;}return true;
}int main(){int T;scanf("%d",&T);while(T--){scanf("%d",&n);for(int i=1; i<=n; ++i)scanf("%lf%lf",&coord[i][0],&coord[i][1]);pos = 0;for(int i=1; i<=n; ++i){for(int j=i+1; j<=n; ++j){arr[pos].u=i, arr[pos].v=j;arr[pos++].val = getDist(coord[i][0],coord[i][1],coord[j][0],coord[j][1]);}}double ans=0;init();sort(arr, arr+pos);for(int i=0; i<pos; ++i){if(Union(arr[i].u, arr[i].v))ans += arr[i].val;}printf("%.2f\n", ans);if(T)puts("");} return 0;
}

2.Prim

#include<cstdio>
#include<cstring>
#include<cmath>
#define N 105
double coord[N][2], w[N][N], minCost[N];
int n, pre[N], hash[N];inline double getDist(double x1,double y1,double x2,double y2){return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); 
}double Prim(){memset(hash, 0, sizeof(hash));hash[1] = 1;for(int i=1; i<=n; ++i){minCost[i] = w[1][i];pre[i] = 1;}double sum=0;for(int i=1; i<n; ++i){int u=-1;for(int j=1; j<=n; ++j)if(!hash[j]){if(u==-1||minCost[j]<minCost[u])u=j;}sum += w[pre[u]][u];hash[u] = 1;for(int j=1; j<=n; ++j)if(!hash[j]){if(minCost[j]>w[u][j]){minCost[j] = w[u][j];pre[j] = u;}}}return sum;
}
int main(){int T;scanf("%d",&T);while(T--){scanf("%d",&n);for(int i=1; i<=n; ++i)scanf("%lf%lf",&coord[i][0],&coord[i][1]);memset(w, 0, sizeof(w));for(int i=1; i<=n; ++i)for(int j=1; j<=n; ++j)if(i!=j)w[i][j] = getDist(coord[i][0],coord[i][1],coord[j][0],coord[j][1]);printf("%.2f\n", Prim());if(T) printf("\n");} return 0;
}

——  生命的意义，在于赋予它意义。
          
     原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)
 

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