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ZOJ T1586 QS Network
题解:
给出了每条边需要的花费,但是本题又加了一个条件,就是每两个点之间要连接电缆的话还得各配一个适配器(一条电缆,两个适配器),每个点都有自己需要的适配器的价格,这就是一条边总的花费。这一点要处理好就OK了。
代码
#include<cstdio>
#include<algorithm>
#define INF 0x3f3f3f3f
#define maxn 1005
using namespace std;
int n,sum,dis[maxn],D[maxn];
int map[maxn][maxn];
bool vis[maxn];
void init(){
sum = 0;
for(int i = 1; i <=n ; i++){
dis[i] = INF;
vis[i] = false;
/*for(int j = 1; j <= i ; j++)
i == j ? map[i][j] = 0 : map[i][j] = map[j][i] = 0;*/
}
}
void Prim(int s){
dis[s] = 0;
for(int i = 1; i <= n ; i++){
int p = 0, minn = INF;
for(int j = 1; j <= n ; j++){
if(!vis[j] && dis[j] < minn){
p = j;
minn = dis[j];
}
}
vis[p] = true;
sum += dis[p];
for(int j = 1; j <= n ; j++){
if(!vis[j] && dis[j] > map[p][j])
dis[j] = map[p][j];
}
}
}
int main(){
int t,num;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
init();
for(int i = 1; i <= n ; i++)
scanf("%d",&D[i]);
for(int i = 1 ; i <= n; i++)
for(int j = 1; j <= n ; j++){
scanf("%d",&map[i][j]);
if(i != j) map[i][j] += D[i] + D[j];
}
Prim(1);
printf("%d\n",sum);
}
return 0 ;
}