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.
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.
本质上就是最小生成树的问题,可以参考这篇文章。我们用Prim实现即可,我们用邻接矩阵,时间复杂度O(V^2),代码如下:
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//POJ 1258 | |
/* | |
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 | |
The 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 lines of 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. | |
Output | |
For 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. | |
*/ | |
#include <stdio.h> | |
#include <string.h> | |
#include <climits> | |
#define INF 0x3f3f3f3f | |
const int N = 105; | |
//minCost[i], curr minimum cost to reach vertex i from visted nodes | |
int graph[N][N],minCost[N],visited[N]; | |
int n; | |
int prim() | |
{ | |
int totalW = 0, currNode = 0; | |
//preprocess, start from node 0 | |
visited[0] = 1; | |
//update all reachable nodes | |
for(int i = 0; i < n; ++i) | |
if(!visited[i]) | |
minCost[i] = graph[currNode][i]; | |
//n - 1 edges to construct | |
for(int i = 1; i < n; ++i) | |
{ | |
//find edge with minimum cost which connects an unvisited node | |
int cost = INT_MAX; | |
for(int j = 0; j < n; ++j) | |
{ | |
if(!visited[j] && minCost[j] < cost) | |
{ | |
cost = minCost[j]; | |
currNode = j; | |
} | |
} | |
//we found the edge | |
visited[currNode] = 1; | |
totalW += cost; | |
//update all reachable nodes | |
for(int k = 0; k < n; ++k) | |
if(!visited[k] && minCost[k] > graph[currNode][k]) | |
minCost[k] = graph[currNode][k]; | |
} | |
return totalW; | |
} | |
int main() | |
{ | |
int i,v,j,ans; | |
while(scanf("%d",&n)!=EOF) | |
{ | |
memset(graph,INF,sizeof(graph)); | |
memset(visited,0,sizeof(visited)); | |
memset(minCost,INF,sizeof(minCost)); | |
for(i = 0; i < n; i++) | |
for(j = 0; j < n; j++) | |
{ | |
scanf("%d",&v); | |
graph[i][j]=graph[i][j]=v; | |
} | |
ans=prim(); | |
printf("%d\n",ans); | |
} | |
return 0; | |
} |
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