HDOJ 1875 畅通工程再续

题意：给出一百个点的坐标，任意两点之间都有边，求将所有点连接起来的最短的所有边长和

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1875

思路：最小生成树模板，稠密图，对边的长度进行筛选即可

注意点：无

以下为AC代码：

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
13162929	2015-03-18 13:37:43	Accepted	1875	171MS	2196K	5231 B	G++	luminous11

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cctype>
#include <climits>
#include <iomanip>
#include <cstdlib>
#include <algorithm>
//#include <unordered_map>
//#include <unordered_set>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#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 RS(s) scanf ( "%s", s )
#define PI(a) printf ( "%d", a )
#define PIL(a) printf ( "%d\n", a )
#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 PL() printf ( "\n" )
#define PSL(s) printf ( "%s\n", s )
#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 )
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);
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 int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };
struct node{
    int x, y, cnt;
    double dis;
    node(){}
    node( int _x, int _y ) : x(_x), y(_y) {}
    node( int _x, int _y, int _cnt ) : x(_x), y(_y), cnt(_cnt) {}
    node( int _x, int _y, double _dis ) : x(_x), y(_y), dis(_dis) {}
    friend bool operator < ( const node &a, const node &b ){
        return a.dis > b.dis;
    }
}p[105];
int n;
double ans;
priority_queue<node> pq;
double calc ( int x, int y )
{
    int xi = p[x].x - p[y].x;
    int yi = p[x].y - p[y].y;
    int tmp = ( xi * xi ) + ( yi * yi );
    return sqrt( (double)tmp );
}
int fa[105];
int ran[105];
int find( int x )
{
    //cout << x << endl;
    return x == fa[x] ? x : find( fa[x] );
}
bool kruskal()
{
    int cnt = 1;
    while ( ! pq.empty() ){
        node tmp = pq.top();
        //cout << tmp.x << ' ' << tmp.y << ' ' << tmp.dis << endl;
        pq.pop();
        int xi = find ( tmp.x );
        int yi = find ( tmp.y );
        //cout << xi << ' ' << yi << endl;
        if ( xi == yi )continue;
        else{
            cnt ++;
            if ( ran[xi] < ran[yi] ){
                fa[xi] = yi;
                ran[yi] = max ( ran[yi], ran[xi] + 1 );
            }
            else{
                fa[yi] = xi;
                ran[xi] = max ( ran[xi], ran[yi] + 1 );
            }
            ans += tmp.dis;
        }
        if ( cnt == n )return true;
    }
    return false;
}
void init()
{
    while ( ! pq.empty() ) pq.pop();
    ans = 0.0;
    rep ( i, 0, 105 )fa[i] = i;
    clr ( ran, 0 );
}
int main()
{
    int T;
    RDI ( T );
    while ( T -- ){
        init();
        RDI ( n );
        rep ( i, 0, n ){
            RDII ( p[i].x, p[i].y );
        }
        double tmp;
        ans = 0.0;
        rep ( i, 0, n ){
            rep ( j, 0, i ){
                tmp = calc ( i, j );
                if ( tmp > 9.9999999 && tmp < 1000.000001 ){
                    pq.push ( node ( i, j, tmp ) );
                }
            }
        }
        if ( kruskal() ){
            printf ( "%0.1f\n", ans * 100 );
        }
        else{
            printf ( "oh!\n" );
        }
    }
    return 0;
}