POJ 1258 Agri-Net(Prim)
阿新 • • 發佈:2017-08-01
rms exc name win style self mar enter needed
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
The distance between any two farms will not exceed 100,000.
題目網址:http://poj.org/problem?id=1258
題目:
Agri-NetTime Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 60004 | Accepted: 24855 |
Description
Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
The distance between any two farms will not exceed 100,000.
Input
Output
Sample Input
4 0 4 9 21 4 0 8 17 9 8 0 16 21 17 16 0
Sample Output
28
最小生成樹水題~因為點數比較少,而且題目給的是鄰接矩陣 所以用了Prim算法。
代碼:
1 #include <cstdio> 2 #include <cstring> 3#include <algorithm> 4 using namespace std; 5 const int inf=11111111; 6 int n,res; 7 int dist[105],f[105][105]; 8 int vis[105]; 9 void prim(){ 10 int v=0; 11 memset(vis, 0, sizeof(vis)); 12 for (int i=1; i<=n; i++) dist[i]=inf; 13 dist[1]=0; 14 for (int i=1; i<=n; i++) { 15 int Min=inf; 16 for (int j=1; j<=n; j++) { 17 if(dist[j]<Min && !vis[j]){ 18 Min=dist[j]; 19 v=j; 20 } 21 } 22 vis[v]=1; 23 res+=dist[v]; 24 for (int j=1; j<=n; j++) { 25 dist[j]=min(dist[j],f[v][j]);//算法核心,是求遍歷點到部分最小生成樹的最小距離,而不是單個點 26 } 27 } 28 } 29 int main(){ 30 while (scanf("%d",&n)!=EOF && n) { 31 res=0; 32 for (int i=1; i<=n; i++) { 33 for (int j=1; j<=n; j++) { 34 scanf("%d",&f[i][j]); 35 } 36 } 37 prim(); 38 printf("%d\n",res); 39 } 40 return 0; 41 }
POJ 1258 Agri-Net(Prim)