I - QS Network

最新推荐文章于 2024-08-12 16:56:06 发布

YRZ-James

最新推荐文章于 2024-08-12 16:56:06 发布

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分类专栏：最小生成树

本文链接：https://blog.youkuaiyun.com/qq_41998938/article/details/86713443

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本文介绍了一道关于QS生物间网络连接的算法题目，通过最小生成树算法解决网络搭建成本最小化问题。考虑到每只QS生物有其偏好的网络适配器品牌，且两两之间的距离不等，任务是确定构建网络的最低成本。

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题目:
Sunny Cup 2003 - Preliminary Round
April 20th, 12:00 - 17:00
Problem E: QS Network

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

代码如下：

#include<iostream>
#include<algorithm>
using namespace std;
#define MAX 1005
struct Node{int x,y,weight;}node[MAX * MAX];
int f[MAX];
int cost[MAX];
int find_father(int x)
{
    int r = x,temp;
    while(r != f[r]) r = f[r];
    while(x != f[x]){
        temp = f[x];
        f[x] = r;
        x = temp;
    }
    return x;
}
void combine(int a,int b)
{
    int c = find_father(a);
    int d = find_father(b);
    if(c == d) return;
    f[d] = c;
}
int kruskal(int m)
{
    int ans = 0;
    for(int i = 0;i < m;i++){
        if(find_father(node[i].x) != find_father(node[i].y)){
            combine(node[i].x,node[i].y);
            ans += node[i].weight;
        }
    }
    return ans;
}
bool cmp(Node a,Node b) {return a.weight < b.weight;}
int main()
{
    int t,n,distance,k,ans;
    cin >> t;
    while(t--){
        cin >> n;
        for(int i = 0;i < n;i++){
            cin >> cost[i];
            f[i] = i;
        }
        k = 0;
        for(int i = 0;i < n;i++){
            for(int j = 0;j < n;j++){
                cin >> distance;
                node[k].x = i;
                node[k].y = j;
                node[k++].weight = distance + cost[i] + cost[j];
            }
        }
        sort(node,node + k,cmp);
        ans = kruskal(k);
        cout << ans << endl;
    }
    return 0;
}

这道题是一道最小生成树的题，与以往题不同，这道题中的权值 = 边的权值 + 两个点的值。之后使用kruskal算法即可。