Building Roads(Prim+Kruskal)

Building Roads
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 11767 Accepted: 3351
Description

Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms.

Each of the N (1 ≤ N ≤ 1,000) farms (conveniently numbered 1..N) is represented by a position (Xi, Yi) on the plane (0 ≤ Xi ≤ 1,000,000; 0 ≤ Yi ≤ 1,000,000). Given the preexisting M roads (1 ≤ M ≤ 1,000) as pairs of connected farms, help Farmer John determine the smallest length of additional roads he must build to connect all his farms.

Input

• Line 1: Two space-separated integers: N and M
• Lines 2..N+1: Two space-separated integers: Xi and Yi
• Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farm i and farm j.

Output

• Line 1: Smallest length of additional roads required to connect all farms, printed without rounding to two decimal places. Be sure to calculate distances as 64-bit floating point numbers.

Sample Input

4 1
1 1
3 1
2 3
4 3
1 4
Sample Output

4.00
Source

USACO 2007 December Silver

n个村庄，m条已经修好的道路，接下来n行表示n个村庄的坐标（xi，yi），m行表示两个村庄（a，b）已经修好道路。求n个村庄互相连通需要修建的最短道路。

Prim:

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#define INF 0x3f3f3f3fdouble ma[1211][1211];double dist[1211];int v[1211];int n, m;int x[1211], y[1211];double dis(int i, int j){    double xx = x[i] - x[j];    double yy = y[i] - y[j];    return sqrt(xx*xx+yy*yy);}void Prim(){    memset(v, 0, sizeof(v));    for(int i=0;i<n;i++)        dist[i] = ma[0][i];    v[0] = 1;    int point;    double ans = 0, min;    for(int i=1;i<n;i++)    {        point = i;        min = INF;        for(int j=0;j<n;j++)        {            if(dist[j]<min&&v[j]==0)            {                point = j;                min = dist[j];            }        }        ans += min;        v[point] = 1;        for(int j=0;j<n;j++)        {            if(dist[j]>ma[point][j]&&v[j]==0)                dist[j] = ma[point][j];        }    }    printf("%.2f\n", ans);}int main(){    scanf("%d %d", &n, &m);    for(int i=0;i<n;i++)    {        scanf("%d %d", &x[i], &y[i]);    }    for(int i=0;i<n;i++)    {        for(int j=0;j<i;j++)        {            ma[i][j] = ma[j][i] = dis(i, j);        }    }    for(int i=0;i<m;i++)    {        int a, b;        scanf("%d %d", &a, &b);        ma[a-1][b-1] = ma[b-1][a-1] = 0;//已经修好的清零    }    Prim();    return 0;}

Kruskal:

#include <iostream>#include <algorithm>#include <cstring>#include <cmath>#include <cstdio>#define INF 0x3f3f3f3fusing namespace std;double x[121001];double y[121001];struct node{    int x, y;    double data;}que[1000000];int n, m, top, cont;int pre[121001];bool cmp(struct node u, struct node b){    return u.data<b.data;}double dis(int i, int j){    double xx = x[i] - x[j];    double yy = y[i] - y[j];    return sqrt(xx*xx+yy*yy);}int root(int a){    while(a!=pre[a])        a = pre[a];    return a;}void Kruskal(){    double ans = 0;    for(int i=0;i<=top&&cont!=n-1;i++)    {        int xx = root(que[i].x);        int yy = root(que[i].y);        if(xx!=yy)            {                pre[xx] = yy;                ans += que[i].data;                cont++;//记录边数            }    }    printf("%.2f\n", ans);}int main(){    cin>>n>>m;    for(int i=1;i<=n;i++)     cin>>x[i]>>y[i];    top = -1, cont = 0;    for(int i=1;i<=n;i++)    {        for(int j=i+1;j<=n;j++)        {            top++;            que[top].x = i;            que[top].y = j;            que[top].data = dis(i, j);        }    }    for(int i=1;i<=n;i++)        pre[i] = i;    for(int i=0;i<m;i++)    {        int a, b;        scanf("%d %d", &a, &b);        int xx = root(a);        int yy = root(b);        if(xx!=yy)            {                pre[xx] = yy;                cont++;            }    }    sort(que, que+top+1, cmp);    Kruskal();    return 0;}