POJ 3265 Building Roads(最小生成树)

Building Roads

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9255		Accepted: 2669

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 (X_i, Y_i) on the plane (0 ≤ X_i≤ 1,000,000; 0 ≤ Y_i≤ 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: X_iand Y_i
* 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

Sample Output

4.00

用Prime算法会比较方便

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <string>
 5 #include <cmath>
 6 #include <iomanip>
 7 using namespace std;
 8 #define maxn 1010
 9 double mp[maxn][maxn];
10 int N, M;
11 int vis[maxn];
12 double low[maxn];
13 struct Point{
14     double x,y;
15 }p[maxn];
16 #define INF 99999999
17 double cal(double x1, double y1, double x2, double y2){
18     return ( sqrt( (x1-x2)*(x1-x2) +(y1-y2)*(y1-y2)));
19 }
20 double prime(){
21     double result;int pos;
22     memset(vis, 0, sizeof(vis));
23     pos = 1; vis[1] = 1; low[1] = 0;
24     for(int i = 1; i <= N; i++){
25         if(i != pos ) low[i] = mp[1][i];  
26     }
27     for(int i = 1; i <= N-1; i++){
28         double min = INF;
29         for(int j = 1; j <= N; j++){
30             if(low[j] < min && !vis[j]){
31                 min = low[j];
32                 pos = j;
33             }
34         }
35         result += min;
36         vis[pos] = 1;
37         
38         for(int j = 1; j <= N; j++){
39             if(!vis[j] && mp[pos][j] < low[j]) low[j] = mp[pos][j];
40         }
41     }
42     return result;
43     
44 }
45 int main(){
46     while(~scanf("%d%d", &N, &M)){
47         memset(mp, 0, sizeof(mp));
48         for(int i = 1; i <= N; i++){
49             scanf("%lf%lf", &p[i].x, &p[i].y);
50         }
51         for(int i = 1; i <= N; i++){
52             for(int j = 1; j <= N; j++){
53                 if(i != j){
54                     mp[i][j] = mp[j][i] = cal(p[i].x, p[i].y,p[j].x, p[j].y);
55                 }
56             }
57         }
58         for(int i = 1; i <= M; i++){
59             int t1, t2;
60             scanf("%d%d", &t1, &t2);
61             mp[t1][t2] = mp[t2][t1] = 0;
62         }
63         cout<<fixed<<setprecision(2)<<prime()<<endl;
64     //    printf("%lf
", prime());
65     }
66     
67     return 0;
68 }