原題鏈接 http://poj.org/problem?id=1258
Description
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
<span style="font-size:24px;color:#333399;">Source Code</span>
/*
最小生成樹 Prim
*/
#include <iostream>
using namespace std;
#define Max 102
#define INF 1000000;
int map[Max][Max];
int prim(int points) {
int i, j, k, MaxDis, result;
bool searched[Max];
int dis[Max];
//初始化,以0爲起點
result = 0;
memset(searched, 0, sizeof(searched));
for (i = 1; i < points; i++) {
dis[i] = map[0][i];
}
searched[0] = 1;
k = 0;
for (i = 1; i < points; i++) {
MaxDis = INF;
//每次找出最近的點
for (j = 0; j < points; j++) {
if (!searched[j] && MaxDis > dis[j]) {
MaxDis = dis[j];
k = j;
}
}
//更新當前點集到其餘未加入的點的最短距離
searched[k] = 1;
result += MaxDis;
for (j = 0; j < points; j++) {
if (!searched[j] && map[k][j] < dis[j]) {
dis[j] = map[k][j];
}
}
}
return result;
}
int main() {
int i, j, farms;
while (scanf("%d", &farms) != EOF) {
for (i = 0; i < farms; i++) {
for (j = 0; j < farms; j++) {
scanf("%d", &map[i][j]);
}
}
printf("%d\n", prim(farms));
}
}