Zoj1586--qs Network (minimum spanning tree)

QS Network

Description
In the planet w-503 of Galaxy CGB, there are a kind of intelligent creature named QS. Qscommunicate with all other via networks. If the QS want to get connected, the they need to buy the network adapters (one for each QS) and a segment of the network cable. Please be advised that one NETWORK ADAPTER CAN is only used in A and 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 their message to all the QS it was connected with, the group of QS Eive the message broadcast the message to all the "Qs they connected with," the procedure repeats until all the QS ' s has a re ceived 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 have its Favorate brand's 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's 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 the data sets.
From the second line there is 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.
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 is non-negative and not more than 1000.
Output
For each data set,output, the minimum cost of 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

Main topic:

QS is an alien (fog?) ), they need a network cable and router to contact (fog?). ）。 Each of them has its own favorite router, and only one cable can be connected to the router (a router can only contact one person).

The input t represents several sets of data. The first behavior of each group of data n, indicates how many QS people, the second behavior of this n personal like the price of the router.

The next n behavior is a n*n matrix that represents the price (distance) of each of them to contact the desired network cable.

Output: The minimum cost of keeping everyone connected.

The idea of knot problem:

The minimum spanning tree. It is important to note that the weight of each edge is not only the price of the network cable, but also the price of the router that includes two vertices.

Code:

1#include <stdio.h>2#include <iostream>3#include <algorithm>4#include <string>5 #defineMAXN 1010*1010/26 using namespacestd;7 structEdge8 {9     intbegin;Ten     intend; One     intdis; A} t[maxn+Ten]; - intfather[maxn+Ten],a[maxn+Ten]; - voidInit () the { -      for(intI=1; i<=maxn;i++) -father[i]=i; - } + intFindintx) - { +     if(father[x]!=x) Afather[x]=find (Father[x]); at     returnFather[x]; - } - voidJoinintXinty) - { -     intFx=find (x), fy=find (y); -     if(fx!=fy) infather[fx]=fy; - } to BOOLcmpstructEdge A,structEdge B) + { -     returna.dis<B.dis; the } * intMain () $ {Panax Notoginseng     intC; -Cin>>C; the      while(c--) +     { A init (); the         intk=1; +         intN; -Cin>>N; $         intCnt=0; $          for(intI=1; i<=n; i++) -Cin>>A[i]; -          for(intI=1; i<=n; i++) the         { -              for(intj=1; j<i; J + +)Wuyi             { theT[k].begin=i,t[k].end=J; -Cin>>T[k].dis; Wut[k].dis+=a[i]+A[j]; -k++; About             } $              for(intj=i;j<=n;j++) -             { -                 inttmp; -Cin>>tmp; A             } +         } theSort (t+1, t+k,cmp); -         intsum=0; $          for(intI=1; i<k;i++) the         { the             //printf ("%d", T[i].dis); the             if(Find (t[i].begin)! =find (t[i].end)) the             { -sum++; incnt+=T[i].dis; the join (t[i].begin,t[i].end); the                 if(sum==n-1) Break; About             } the         } theprintf"%d\n", CNT); the     } +     return 0; -}