POJ 2349 Arctic Network

题意：在若干个村庄之间需要有卫星天线和无线发信器，卫星天线之间的通讯可以无视距离，无线发信器之间通讯有距离限制，已知每个村庄的坐标，已经卫星天线的个数以及村庄的个数，求无线发信器的通讯距离至少要多少

链接：http://poj.org/problem?id=2349

思路：求最小生成树上的第k大边权，建图后利用模板求出最小生成树，利用dis数组的性质求出第k大边权

注意点：无

以下为AC代码：

Run IDUserProblemResultMemoryTimeLanguageCode LengthSubmit Time14176789luminous112349Accepted2724K47MSG++4433B2015-05-09 09:53:55

/* ************************************************# @Author  : Luminous11 ([e-mail])*# @Date    : 2015-04-20 20:15:06*# @Link    : http://blog.csdn.net/luminous11                          *********************************************** */#include <algorithm>#include <iostream>#include <climits>#include <iomanip>#include <cstdlib>#include <cstring>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <deque>#include <list>#include <map>#include <set>//#include <unordered_map>//#include <unordered_set>#define pb push_back#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define READ(f) freopen(f, "r", stdin)#define WRITE(f) freopen(f, "w", stdout)#define clr(a, v) memset( a , v , sizeof(a) )#define RS(s) scanf ( "%s", s )#define RDI(a) scanf ( "%d", &a )#define RDII(a, b) scanf ( "%d%d", &a, &b )#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c )#define PL() printf ( "\n" )#define PI(a) printf ( "%d", a )#define PIL(a) printf ( "%d\n", a )#define PSL(s) printf ( "%s\n", s )#define PII(a,b) printf ( "%d %d", a, b )#define PIIL(a,b) printf ( "%d %d\n", a, b )#define PIII(a,b,c) printf ( "%d %d %d", a, b, c )#define PIIIL(a,b,c) printf ( "%d %d %d\n", a, b, c )#define rep(i,m,n) for ( int i = m; i <  n; i ++ )#define REP(i,m,n) for ( int i = m; i <= n; i ++ )#define dep(i,m,n) for ( int i = m; i >  n; i -- )#define DEP(i,m,n) for ( int i = m; i >= n; i -- )#define repi(i,m,n,k) for ( int i = m; i <  n; i += k )#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )#define depi(i,m,n,k) for ( int i = m; i >  n; i += k )#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )//#pragma comment ( linker, "/STACK:1024000000,1024000000" )using namespace std;template <class T>inline bool RD ( T &ret ){    char c;    int sgn;    if ( c = getchar(), c ==EOF )return 0; //EOF    while ( c != '-' && ( c < '0' || c > '9' ) ) c = getchar();    sgn = ( c == '-' ) ? -1 : 1;    ret = ( c == '-' ) ? 0 : ( c - '0' );    while ( c = getchar() , c >= '0' && c <= '9' ) ret = ret * 10 + ( c - '0' );    ret *= sgn;    return 1;}inline void PD ( int x ){    if ( x > 9 ) PD ( x / 10 );    putchar ( x % 10 + '0' );}const double eps = 1e-10;const double pi = acos(-1.0);const int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };struct node{    int x, y, cnt;    node(){}    node( int _x, int _y ) : x(_x), y(_y) {}    node( int _x, int _y, int _cnt ) : x(_x), y(_y), cnt(_cnt) {}};struct edge{    int u, v;    double wei;    edge(){}    edge( int _u, int _v, double _w ) : u(_u), v(_v), wei(_w) {}};bool operator< ( const edge & a, const edge &b ){    return a.wei < b.wei;}int s, p;double ans;node num[505];double g[505][505];double dis[505];bool vis[505];double calc ( int u, int v ){    int x = num[u].x - num[v].x;    int y = num[u].y - num[v].y;    double dis = sqrt ( x * x + y * y );    return dis;}void init(){    clr ( num, 0 );    clr ( vis, 0 );    rep ( i, 0, 505 ){        dis[i] = 0.0;        rep ( j, 0, 505 ){            g[i][j] = 1000000000;        }    }}void prim(){    REP ( i, 1, p ){        dis[i] = g[1][i];    }    dis[1] = 0.0;    rep ( k, 0, p ){        double _min = 100000000.0;        int l = -1;        REP ( i, 1, p ){            if ( ! vis[i] && _min > dis[i] ){                _min = dis[i];                l = i;            }        }        vis[l] = 1;        REP ( i, 1, p ){            if ( ! vis[i] && dis[i] > g[l][i] ){                dis[i] = g[l][i];            }        }    }}bool cmp ( const int &a, const int &b ){    return a > b;}int main(){    int T;    RDI ( T );    while ( T -- ){        init();        RDII ( s, p );        int v, u, w;        REP ( i, 1, p ){            RDII ( u, v );            num[i] = node ( u, v );        }        int cnt = 0;        REP ( i, 1, p ){            REP ( j, i + 1, p ){                g[i][j] = g[j][i] = calc ( i, j );            }        }        prim();        sort ( dis, dis + p + 1, cmp );        cout << fixed << setprecision(2) << dis[s-1] << endl;    }    return 0;}