FOJ 1096 QS Network (最小生成树kruskal)

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本文介绍了一种基于最小生成树算法解决智能生物QS通信网络构建的问题。通过计算不同QS间的电缆成本及网络适配器费用，利用Kruskal算法确定最低成本的网络连接方案。

Problem Description
In the planet w-503 of galaxy cgb, there is a kind of intelligent creature named QS. QScommunicate with each other via networks. If two QS want to get connected, they need to buy two network adapters (one for each QS) and a segment of network cable. Please be advised that ONE NETWORK ADAPTER CAN ONLY BE USED IN A SINGLE CONNECTION.(ie. if a QS want to setup four connections, it needs to buy four adapters). In the procedure of communication, a QS broadcasts its message to all the QS it is connected with, the group of QS who receive the message broadcast the message to all the QS they connected with, the procedure repeats until all the QS's have received the message.
A sample is shown below:
A sample QS network, and QS A want to send a message.
Step 1. QS A sends message to QS B and QS C;
Step 2. QS B sends message to QS A ; QS C sends message to QS A and QS D;
Step 3. the procedure terminates because all the QS received the message.
Each QS has its favorate brand of network adapters and always buys the brand in all of its connections. Also the distance between QS vary. Given the price of each QS's favorate brand of network adapters and the price of cable between each pair of QS, your task is to write a program to determine the minimum cost to setup a QS network.
Input
The 1st line of the input contains an integer t which indicates the number of data sets.
From the second line there are t data sets.
In a single data set,the 1st line contains an interger n which indicates the number of QS.
The 2nd line contains n integers, indicating the price of each QS's favorate network adapter.
In the 3rd line to the n+2th line contain a matrix indicating the price of cable between ecah pair of QS.
Constrains:
All the integers in the input are non-negative and not more than 1000.
Output
for each data set,output the minimum cost in a line. NO extra empty lines needed.
Sample Input
1
3
10 20 30
0 100 200
100 0 300
200 300 0
Sample Output

370

给你n个点，为了通信，就是找出n-1条边连接起来，要求是你所得到的花费最小。这边的花费代表的是两点间电缆的钱加上通信设备的钱；假如a与b这两个点之间要进行通信，那么这两点之间的花费就是a, b电缆的钱加上a， b两点各买通讯设备的钱，不同的两点要有不同的通讯设备，所以开始先一遍把两两之间的花费求出来，然后就是赤裸裸的最小生成树。

AC代码：

# include <stdio.h>
# include <algorithm>
using namespace std;
int cable[1010][1010];
int a[1010];
int father[1010];
int n;
void init(){//初始化根节点为本身
	for(int i=1; i<=n; i++)
	father[i]=i;
}
int find_father(int n){//并查集
	if(n==father[n]){
		return n;
	}
	return father[n]=find_father(father[n]);
}
struct side{
	int begin, end;
	int l;
};
int  cmp(side a, side b){
	return a.l<b.l;
}
side s[500010];
int main(){
	int i, j, k, t, m, cost;
	scanf("%d", &t);
	while(t--){
		scanf("%d", &n);
		for(i=1; i<=n; i++){
			scanf("%d", &a[i]);
		}
		for(i=1; i<=n; i++){
			for(j=1; j<=n; j++){
				scanf("%d", &cable[i][j]);
			}
		}
		init();
		m=0;
		for(i=1; i<=n-1; i++){//求两两之间花费
			for(j=i+1; j<=n; j++){
				s[m].begin=i;
				s[m].end=j;
				s[m].l=cable[i][j]+a[i]+a[j];
				m++;
			}
		}
		sort(s, s+m, cmp);
		cost=0;
		for(i=0; i<=m-1; i++){//kruskal算法
			int f1=find_father(s[i].begin);
			int f2=find_father(s[i].end);
			if(f1!=f2){
				cost=s[i].l+cost;
				father[f1]=f2;
			}
		}
		printf("%d\n", cost);
	}
	return 0;
}