﻿ zoj 1203 Swordfish

### zoj 1203 Swordfish

```  1 /**************************************************************************
2 user_id: SCNU20102200088
3 problem_id: zoj 1203
4 problem_name: Swordfish
5 **************************************************************************/
6
7 #include <algorithm>
8 #include <iostream>
9 #include <iterator>
10 #include <iomanip>
11 #include <sstream>
12 #include <fstream>
13 #include <cstring>
14 #include <cstdlib>
15 #include <climits>
16 #include <bitset>
17 #include <string>
18 #include <vector>
19 #include <cstdio>
20 #include <cctype>
21 #include <ctime>
22 #include <cmath>
23 #include <queue>
24 #include <stack>
25 #include <list>
26 #include <set>
27 #include <map>
28 using namespace std;
29
30 //线段树
31 #define lson l,m,rt<<1
32 #define rson m+1,r,rt<<1|1
33
34 //手工扩展栈
35 #pragma comment(linker,"/STACK:102400000,102400000")
36
37 const double EPS=1e-9;
38 const double PI=acos(-1.0);
39 const double E=2.7182818284590452353602874713526;  //自然对数底数
40 const double R=0.5772156649015328606065120900824;  //欧拉常数:(1+1/2+...+1/n)-ln(n)
41
42 const int x4[]={-1,0,1,0};
43 const int y4[]={0,1,0,-1};
44 const int x8[]={-1,-1,0,1,1,1,0,-1};
45 const int y8[]={0,1,1,1,0,-1,-1,-1};
46
47 typedef long long LL;
48
49 typedef int T;
50 T max(T a,T b){ return a>b? a:b; }
51 T min(T a,T b){ return a<b? a:b; }
52 T gcd(T a,T b){ return b==0? a:gcd(b,a%b); }
53 T lcm(T a,T b){ return a/gcd(a,b)*b; }
54
55 ///////////////////////////////////////////////////////////////////////////
56 //Add Code:
57 int n,s[105];
58
59 struct Node{
60     int u,v;
61     double w;
62     bool operator <(const Node &a) const{
63         return w>a.w;
64     }
65 }node;
66
67 priority_queue<Node> pq;
68
69 double dist(double x1,double y1,double x2,double y2){
70     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
71 }
72
73 void Union(int x,int y){
74     s[y]=x;
75 }
76
77 int Find(int x){
78     if(s[x]<0) return x;
79     return s[x]=Find(s[x]);
80 }
81
82 double Kruskal(){
83     int num=0;
84     double ret=0;
85     memset(s,-1,sizeof(s));
86     while(!pq.empty() && num<n-1){
87         node=pq.top();
88         pq.pop();
89         if(Find(node.u)!=Find(node.v)){
90             Union(Find(node.u),Find(node.v));
91             ret+=node.w;
92             num++;
93         }
94     }
95     while(!pq.empty()) pq.pop();
96     return ret;
97 }
98 ///////////////////////////////////////////////////////////////////////////
99
100 int main(){
101     std::ios::sync_with_stdio(false);
102     //freopen("in.txt","r",stdin);
103     //freopen("out.txt","w",stdout);
104     ///////////////////////////////////////////////////////////////////////
105     //Add Code:
106     int Case=1,i,j;
107     double x[105],y[105];
108     while(scanf("%d",&n)!=EOF){
109         if(n==0) break;
110         for(i=1;i<=n;i++) scanf("%lf%lf",&x[i],&y[i]);
111         for(i=1;i<n;i++){
112             for(j=i+1;j<=n;j++){
113                 node.u=i;
114                 node.v=j;
115                 node.w=dist(x[i],y[i],x[j],y[j]);
116                 pq.push(node);
117             }
118         }
119         double ans=Kruskal();
120         if(Case>1) printf("
");
121         printf("Case #%d:
",Case++);
122         printf("The minimal distance is: %.2lf
",ans);
123     }
124     ///////////////////////////////////////////////////////////////////////
125     return 0;
126 }
127
128 /**************************************************************************
129 Testcase:
130 Input:
131 5
132 0 0
133 0 1
134 1 1
135 1 0
136 0.5 0.5
137 0
138 Output:
139 Case #1:
140 The minimal distance is: 2.83
141 **************************************************************************/```