ZOJ 1586 QS Network

Description：

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

题目大意：

给定n个点， 从任意一点向另外一个点连边的代价是点的权值加上连边的权值， 求连通所有点的最小花费。

解题思路：

在最小生成树的基础上只要更改加边时直接将点权加在边上， 这样就避免了在跑Prim时候不断地加点权和减点权。

（提醒一点ZOJ很严， 题目上说所有值都不会超过1000， 数组就老老实实开别开太大。 一开始开了1e4 + ９的二维数组一直返回Segmentation Fault错了好几次。。。）

代码：

#include <iostream>#include <sstream>#include <cstdio>#include <algorithm>#include <cstring>#include <iomanip>#include <utility>#include <string>#include <cmath>#include <vector>#include <bitset>#include <stack>#include <queue>#include <deque>#include <map>#include <set>using namespace std;/*tools: *ios::sync_with_stdio(false); *freopen("input.txt", "r", stdin); */typedef long long ll;typedef unsigned long long ull;const int dir[5][2] = {0, 1, 0, -1, 1, 0, -1, 0, 0, 0};const ll ll_inf = 0x7fffffff;const int inf = 0x3f3f3f;const int mod = 1000000;const int Max = (int) 1e3 + 9;int Map[Max][Max], vis[Max], pri[Max];int n, q;void Init() {    for (int i = 0; i <= n; ++i) {        vis[i] = 0; pri[i] = 0;        for (int j = 0; j <= n; ++j) {            Map[i][j] = (i == j) ? 0 : inf;        }    }}void Prim() {    int dis[Max], ans = 0;    for (int i = 1; i <= n; ++i) {        dis[i] = Map[1][i];    }    dis[1] = 0;    for (int i = 1; i <= n; ++i) {        int Min = inf, mark;        for (int j = 1; j <= n; ++j) {            if (!vis[j] && dis[j] < Min) {                Min = dis[mark = j];            }        }        vis[mark] = 1;        ans += dis[mark];        for (int j = 1; j <= n; ++j) {            if (!vis[j] && dis[j] > Map[mark][j]) {                dis[j] = Map[mark][j];            }        }    }    printf("%d\n", ans);}int main() {    //freopen("input.txt", "r", stdin);    int t;    // initialization    scanf("%d", &t);    while (t--) {        Init();        scanf("%d", &n);        for (int i = 1; i <= n; ++i) {            scanf("%d", &pri[i]);        }        for (int i = 1; i <= n; ++i) {            for (int j = 1; j <= n; ++j) {                scanf("%d", &Map[i][j]);                Map[i][j] += pri[i];                Map[i][j] += pri[j];            }        }        Prim();    }    return 0;}