poj 2560 Freckles(并查集）

Freckles

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8749 Accepted: 4177
Description

In an episode of the Dick Van Dyke show, little Richie connects the freckles on his Dad’s back to form a picture of the Liberty Bell. Alas, one of the freckles turns out to be a scar, so his Ripley’s engagement falls through.
Consider Dick’s back to be a plane with freckles at various (x,y) locations. Your job is to tell Richie how to connect the dots so as to minimize the amount of ink used. Richie connects the dots by drawing straight lines between pairs, possibly lifting the pen between lines. When Richie is done there must be a sequence of connected lines from any freckle to any other freckle.
Input

The first line contains 0 < n <= 100, the number of freckles on Dick’s back. For each freckle, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the freckle.
Output

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the freckles.
Sample Input

3
1.0 1.0
2.0 2.0
2.0 4.0
Sample Output

3.41

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给出一些坐标点，求连通所有点所需最短路径。
可以用最小生成树kruskal算法。

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#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;#define N 111#define inf 0x3f3f3f3fint p[N];struct node{    double x, y, z;}nod[N],way[N*N];bool cmp( node a, node b ){ return a.z < b.z; }int fnd( int x ) { return p[x] = p[x] == x ? x : fnd( p[x] ); }int main(){    //freopen( "in.txt", "r", stdin );    int t, cnt = 1;    double sum = 0;    scanf( "%d", &t );    for( int i = 1 ; i <= t ; i ++ ){        scanf( "%lf%lf", &nod[i].x, &nod[i].y );    }    for( int i = 1 ; i <= t ; i ++ ){        for( int j = 1 ; j < i ; j ++ ){            way[cnt].x = i;            way[cnt].y = j;            way[cnt++].z = sqrt( (nod[i].x-nod[j].x)*(nod[i].x-nod[j].x)+(nod[i].y-nod[j].y)*(nod[i].y-nod[j].y));        }    }    sort( way+1, way+cnt, cmp );///    for( int i = 0 ; i <= t ; i ++ ) p[i] = i;    for( int i = 1 ; i < cnt ; i ++ ){        int u = fnd( way[i].x );        int v = fnd( way[i].y );        if( u != v ){            p[u] = way[i].y;            sum += way[i].z;        }    }    printf( "%.2lf\n", sum );    return 0;}