【ZOJ】1586 - QS Network（克鲁斯塔尔）

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

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Author: JIANG, Jiefeng
Source: Zhejiang University Local Contest 2003, Preliminary

读题：这道题刚开始看很不好理解啊，后来没办法只好借助了电子翻译。题目本来和多数最小生成数的题意一样，给出两个点的价格，但是这道题有点特殊：两点如果需要连接，每个QS都需要一个适配器，而且是每次连接都需要各自配一个适配器。这点读出来就不难了。

这道题WA了两次，因为测试数据就一组，bug找了半天也找不出来。

之前的问题出在用克鲁斯塔尔的时候，保存两个点之间的花费的时候，应该把各自买适配器的价格都加上。

另外，保存数据的时候尽量开大一点。

代码如下：

#include <cstdio>#include <algorithm>using namespace std;int f[11111];int find (int x){if (x!=f[x])f[x]=find(f[x]);return f[x];}int join(int x,int y){int fx,fy;fx=find(x);fy=find(y);if (fx!=fy){f[fx]=fy;return 1;}return 0;}struct node{int st,end,cost;}data[511111];bool cmp(node a,node b){return a.cost<b.cost;}int main(){int u,n;int ans;int t;int p[11111];//自己设备的价格scanf ("%d",&u);while (u--){scanf ("%d",&n);for (int i=1;i<=n;i++)f[i]=i;for (int i=1;i<=n;i++)scanf ("%d",&p[i]);int num=-1;for (int i=1;i<=n;i++){for (int j=1;j<=n;j++){scanf ("%d",&t);if (j>i){num++;data[num].st=i;data[num].end=j;data[num].cost=t+p[i]+p[j];//这里应该注意：加上所需设备的价格，而不是在后面计算的时候加 }}}sort (data,data+num+1,cmp);ans=0;for (int i=0;i<=num;i++){if (join(data[i].st,data[i].end))ans+=data[i].cost;}printf ("%d\n",ans);}return 0;}