zoj 1203 Swordfish

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1203

解题思路：最小生成树（kruskal 算法 + 优先队列 + 并查集）

/**************************************************************************
user_id: SCNU20102200088
problem_id: zoj 1203
problem_name: Swordfish
**************************************************************************/

#include <algorithm>
#include <iostream>
#include <iterator>
#include <iomanip>
#include <sstream>
#include <fstream>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;

//线段树
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

//手工扩展栈
#pragma comment(linker,"/STACK:102400000,102400000")

const double EPS=1e-9;
const double PI=acos(-1.0);
const double E=2.7182818284590452353602874713526;  //自然对数底数
const double R=0.5772156649015328606065120900824;  //欧拉常数:(1+1/2+...+1/n)-ln(n)

const int x4[]={-1,0,1,0};
const int y4[]={0,1,0,-1};
const int x8[]={-1,-1,0,1,1,1,0,-1};
const int y8[]={0,1,1,1,0,-1,-1,-1};

typedef long long LL;

typedef int T;
T max(T a,T b){ return a>b? a:b; }
T min(T a,T b){ return a<b? a:b; }
T gcd(T a,T b){ return b==0? a:gcd(b,a%b); }
T lcm(T a,T b){ return a/gcd(a,b)*b; }

///////////////////////////////////////////////////////////////////////////
//Add Code:
int n,s[105];

struct Node{
    int u,v;
    double w;
    bool operator <(const Node &a) const{
        return w>a.w;
    }
}node;

priority_queue<Node> pq;

double dist(double x1,double y1,double x2,double y2){
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

void Union(int x,int y){
    s[y]=x;
}

int Find(int x){
    if(s[x]<0) return x;
    return s[x]=Find(s[x]);
}

double Kruskal(){
    int num=0;
    double ret=0;
    memset(s,-1,sizeof(s));
    while(!pq.empty() && num<n-1){
        node=pq.top();
        pq.pop();
        if(Find(node.u)!=Find(node.v)){
            Union(Find(node.u),Find(node.v));
            ret+=node.w;
            num++;
        }
    }
    while(!pq.empty()) pq.pop();
    return ret;
}
///////////////////////////////////////////////////////////////////////////

int main(){
    std::ios::sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    ///////////////////////////////////////////////////////////////////////
    //Add Code:
    int Case=1,i,j;
    double x[105],y[105];
    while(scanf("%d",&n)!=EOF){
        if(n==0) break;
        for(i=1;i<=n;i++) scanf("%lf%lf",&x[i],&y[i]);
        for(i=1;i<n;i++){
            for(j=i+1;j<=n;j++){
                node.u=i;
                node.v=j;
                node.w=dist(x[i],y[i],x[j],y[j]);
                pq.push(node);
            }
        }
        double ans=Kruskal();
        if(Case>1) printf("
");
        printf("Case #%d:
",Case++);
        printf("The minimal distance is: %.2lf
",ans);
    }
    ///////////////////////////////////////////////////////////////////////
    return 0;
}

/**************************************************************************
Testcase:
Input:
5
0 0
0 1
1 1
1 0
0.5 0.5
0
Output:
Case #1:
The minimal distance is: 2.83
**************************************************************************/