ZOJ 1586 QS Network(最小生成树)

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

这一题虽然简单，但也是一个非常有意思的题，这个题的题意就是做一个最小生成树，不过没两个点的连接需要消耗一个连接器，比如1这个点连接了2和3，那么1就需要消耗两个连接器，最开始我的做法是用prim跑一下，然后再把连接的边的两边的点消耗加入进去，但WA了，后来想了一下，应该是先把点的消耗也记录到边的权值里面，比如1和2连接话费100，1连接器价格为20，2为30，那么就把1和2连接的话费变为150，再用一下prim就可以跑出结果了

代码如下:

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;
const int MAX_V = 1100;
const int INF = 0x3f3f3f3f;
int cost[MAX_V],e[MAX_V][MAX_V];
int extra[MAX_V];
bool used[MAX_V];
int V;
int prim(int a_){
    int res = 0;
    fill(cost+1,cost+1+V,INF);
    memset(used,false,sizeof(used));
    cost[a_] = 0;
    while(true){
        int u = -1;
        for(int i=1;i<=V;i++){
            if(!used[i] && (u == -1 || cost[i] < cost[u]))
                u = i;
        }
        if(u == -1) break;
        used[u] = true;
        res += cost[u];
        for(int i=1;i<=V;i++){
            if(!used[i] && cost[i] > e[u][i]){
                cost[i] = e[u][i];
            }
        }
    }
    return res;
}
int main(void){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&V);
        for(int i=1;i<=V;i++)
            scanf("%d",&extra[i]);
        for(int i=1;i<=V;i++){
            for(int j=1;j<=V;j++){
                scanf("%d",&e[i][j]);
            }
        }
        for(int i=1;i<=V;i++)
            for(int j=i+1;j<=V;j++){
                e[i][j] += extra[i] + extra[j];
                e[j][i] = e[i][j];
            }
        printf("%d\n",prim(1));
    }

    return 0;
}