ZOJ1586——QS Network(最小生成树)
QS Network
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
题目大意：
QS是一种外星人（雾？），他们之间需要网线和路由器来联系（大雾？）。他们每个人都有自己喜欢的路由器，并且路由器上只能接一条网线（一个路由器只能与一个人联系）。
输入T代表几组数据。每组数据的第一行为N，表示有多少个QS人，第二行为这N个人喜欢的路由器的价格。
后面的N行为一个N*N的矩阵，表示他们之间各自的要想联系需要的网线的价格（距离）。
输出：保证所有人相连的最小花销。
结题思路：
最小生成树。需要注意的是 每条边的权值不单单是网线的价格还要包括两个顶点的路由器价格。
Code：
1 #include<stdio.h> 2 #include<iostream> 3 #include<algorithm> 4 #include<string> 5 #define MAXN 1010*1010/2 6 using namespace std; 7 struct edge 8 { 9 int begin; 10 int end; 11 int dis; 12 } T[MAXN+10]; 13 int father[MAXN+10],a[MAXN+10]; 14 void init() 15 { 16 for (int i=1;i<=MAXN;i++) 17 father[i]=i; 18 } 19 int find(int x) 20 { 21 if (father[x]!=x) 22 father[x]=find(father[x]); 23 return father[x]; 24 } 25 void join(int x,int y) 26 { 27 int fx=find(x),fy=find(y); 28 if (fx!=fy) 29 father[fx]=fy; 30 } 31 bool cmp(struct edge a,struct edge b) 32 { 33 return a.dis<b.dis; 34 } 35 int main() 36 { 37 int C; 38 cin>>C; 39 while (C--) 40 { 41 init(); 42 int k=1; 43 int N; 44 cin>>N; 45 int cnt=0; 46 for (int i=1; i<=N; i++) 47 cin>>a[i]; 48 for (int i=1; i<=N; i++) 49 { 50 for (int j=1; j<i; j++) 51 { 52 T[k].begin=i,T[k].end=j; 53 cin>>T[k].dis; 54 T[k].dis+=a[i]+a[j]; 55 k++; 56 } 57 for (int j=i;j<=N;j++) 58 { 59 int tmp; 60 cin>>tmp; 61 } 62 } 63 sort(T+1,T+k,cmp); 64 int sum=0; 65 for (int i=1;i<k;i++) 66 { 67 //printf("%d ",T[i].dis); 68 if (find(T[i].begin)!=find(T[i].end)) 69 { 70 sum++; 71 cnt+=T[i].dis; 72 join(T[i].begin,T[i].end); 73 if (sum==N-1) break; 74 } 75 } 76 printf("%d\n",cnt); 77 } 78 return 0; 79 }