POJ1258 并查集和最小生成树

Agri-Net

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 47782		Accepted: 19806

DescriptionFarmer 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.InputThe input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N linesof N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.OutputFor each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.Sample Input

4
0 4 9 21
4 0 8 17
9 8 0 16
21 17 16 0

Sample Output

28

题目大意：总统要通过打电话的形式联系所有农民，然后要求铺电话线的线路最短，该怎么铺遇到的问题和思路：表示最近这类问题做的有点多。。。就不多说了。。。给出prim算法：

#include#include#include#includeusing namespace std;const int inf = 0x3f3f3f3f;int n;int cost[150][150];int mincost[150];bool used[150];void solve(){memset(used, 0, sizeof(used));mincost[1] = 0;int res = 0;while(true){int v = -1;for(int i = 1; i <= n; i++){if(!used[i] && (v == -1 || mincost[i] < mincost[v])) v = i;}if(v == -1)break;res += mincost[v];used[v] = true;for(int i = 1; i <= n; i++){mincost[i] = min(mincost[i], cost[v][i]);}}printf("%d\n", res);}int main(){while(scanf("%d", &n)!=EOF){for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){scanf("%d", &cost[i][j]);}}fill(mincost + 1, mincost + 1 + n, inf);solve();}return 0;}