ZOJ1586 QS Network

一.原题链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=586

二.题目大意：说了那么多，就是给一个邻接矩阵，让你求最短路，注意要边的权要先加上2个点的适配器价格之和。

三.代码：

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

using namespace std;

const int MAX_SIZE = 1002,
INF = 0x3f3f3f3f,
MOD = 1000000007;

int nodeNum, nodeCost[MAX_SIZE],
graph[MAX_SIZE][MAX_SIZE],
lowCost[MAX_SIZE];

int Prim(int start)
{
int i, j, minEdge, res, v;
for(i = 0; i < nodeNum; i++)
lowCost[i] = graph[start][i];

lowCost[start] = -1;

res = 0;
for(i = 1; i < nodeNum; i++){

minEdge = INF;

for(j = 0; j < nodeNum; j++)
if(lowCost[j] != -1 && lowCost[j] < minEdge){
minEdge = lowCost[j];
v = j;
}

lowCost[v] = -1;
res += minEdge;

for(j = 0; j < nodeNum; j++)
if(lowCost[j] > graph[v][j])
lowCost[j] = graph[v][j];
}

return res;
}

int main()
{
//freopen("in.txt", "r", stdin);
int i, j, test;

cin>>test;
while(test--){

cin>>nodeNum;

for(i = 0; i < nodeNum; i++)
cin>>nodeCost[i];

for(i = 0; i < nodeNum; i++)
for(j = 0; j < nodeNum; j++){
cin>>graph[i][j];
graph[i][j] += nodeCost[i] + nodeCost[j];
}

cout<<Prim(0)<<endl;
}
}