最小生成树（prim和Kruskal操！！SB题）

Arctic Network

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 30571		Accepted: 9220

Description

The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver and some outposts will in addition have a satellite channel.
Any two outposts with a satellite channel can communicate via the
satellite, regardless of their location. Otherwise, two outposts can
communicate by radio only if the distance between them does not exceed
D, which depends of the power of the transceivers. Higher power yields
higher D but costs more. Due to purchasing and maintenance
considerations, the transceivers at the outposts must be identical; that
is, the value of D is the same for every pair of outposts.

Your job is to determine the minimum D required for the
transceivers. There must be at least one communication path (direct or
indirect) between every pair of outposts.

Input

The
first line of input contains N, the number of test cases. The first
line of each test case contains 1 <= S <= 100, the number of
satellite channels, and S < P <= 500, the number of outposts. P
lines follow, giving the (x,y) coordinates of each outpost in km
(coordinates are integers between 0 and 10,000).

Output

For
each case, output should consist of a single line giving the minimum D
required to connect the network. Output should be specified to 2 decimal
points.

Sample Input

Sample Output

212.13

Source

Waterloo local 2002.09.28

这个sb题要注意几点：第一是memset对double类型ma初始化为INF时会出现问题。

第二点：最后输出不知道为什么这个sb题得输出非得时%.2f，%.2lf就一直wa。。。。

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <iostream>

#include <algorithm>

#include <iostream>

#include<cstdio>

#include<string>

#include<cstring>

#include <stdio.h>

#include <queue>

#include <string.h>

#include <vector>

#include <map>

#define ME(x , y) memset(x , y , sizeof(x))

#define SF(n) scanf("%d" , &n)

#define rep(i , n) for(int i = 0 ; i < n ; i ++)

#define INF  0x3f3f3f3f

#define mod 1024

using namespace std;

typedef long long ll ;

int s , n ;

double ma[][];

double dis[] ;

double d[];

int  vis[] ;

int ans ;

struct node{

    double x , y ;

}a[];

void prim()

{

    for(int i =  ; i <= n ; i++)

    {

        dis[i] = ma[][i];

    }

    vis[] =  ;

    for(int i =  ; i < n ; i++)

    {

        double min = INF ;

        int pos ;

        for(int j =  ; j <= n ; j++)

        {

            if(!vis[j] && dis[j] < min)

            {

                min = dis[j];

                pos = j ;

            }

        }

        vis[pos] =  ;

        d[ans++] = min ;

        for(int j =  ; j <= n ; j++)

        {

            if(!vis[j] && dis[j] > ma[pos][j])

            {

                dis[j] = ma[pos][j];

            }

        }

    }

}

bool cmp(double a , double b)

{

    return a > b ;

}

void init()

{

    memset(vis ,  , sizeof(vis));

    for(int i =  ; i <= n ; i++)

    {

        for(int j =  ; j <= n ; j++)

        {

            ma[i][j] = INF;

        }

    }

    ans =  ;

}

int main()

{

    int t ;

    scanf("%d" , &t);

    while(t--)

    {

        scanf("%d%d" , &s , &n);

        init();

        for(int i =  ; i <= n ; i++)

        {

            scanf("%lf%lf" , &a[i].x , &a[i].y);

        }

        for(int i =  ; i <= n ; i++)

        {

            for(int j = i +  ; j <= n ; j++)

            {

                double w ;

                w = sqrt((a[i].y - a[j].y) * (a[i].y - a[j].y) + (a[i].x - a[j].x)*(a[i].x - a[j].x)) ;

                ma[i][j] = ma[j][i] = min(ma[i][j] , w);//这里if判断大小时，注意方向不要搞错了

            }

        }

        prim();

        sort(d , d + n -  , cmp);

        printf("%.2f\n" , d[s-]);

    }

    return  ;

}

Kruskal

#include <string.h>

#include <iostream>

#include <stdio.h>

#include <algorithm>

#include <math.h>

using namespace std;

int  fa[] , ans  , n , m  , way ;

double ds[] ;

struct Node

{

    int from , to ;

    int x ;

    int y ;

    double  d;

}a[];

bool cmp(const Node &a , const Node &b)

{

    return a.d < b.d ;

}

int gcd(int x)

{

    if(x == fa[x])

        return x ;

    else

        return gcd(fa[x]);

}

void unite(int x , int y)

{

    x = gcd(x) ; y = gcd(y) ;

    if(x > y) fa[x] = y ;

    else fa[y] = x ;

}

void init()

{

    for(int i =  ; i <= m  ; i++)

        fa[i] = i ;

        ans =  ;

        way =  ;

        memset(ds ,  , sizeof(ds));

}

bool cmp1(double a , double b)

{

    return a > b ;

}

int main()

{

    int t ;

    cin >> t ;

    while(t--)

    {

        cin >> n >> m ;

        init();

        for(int i =  ; i <= m ; i++)

        {

            cin >> a[i].x >> a[i].y ;

        }

        for(int i =  ; i <= m ; i++)

        {

            for(int j = i +  ; j <= m ; j++)

            {

                a[way].from = i ;

                a[way].to = j ;

                a[way].d = sqrt(pow((double)(a[i].x - a[j].x) , ) + pow((double)(a[i].y - a[j].y) , ));

                way++;

            }

        }

        sort(a , a + way , cmp) ;

        for(int i =  ; i < way ; i++)

        {

            if( ans == m - )

                break ;

            if(gcd(fa[a[i].from]) != gcd(fa[a[i].to]))

            {

                unite(a[i].from , a[i].to);

                ds[ans] = a[i].d;

                ans ++ ;

            }

        }

        sort(ds , ds + ans , cmp1);

        printf("%.2f\n" , ds[n-]);

    }

    return ;

}

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最小生成树（prim和Kruskal操！！SB题）

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