| 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 (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
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 }