B - Networking——最小生成树_Prim算法(未堆优化)

Think:

1知识点：最小生成树_Prim算法(未堆优化)

2题意分析：

1>n个点，选择最优方案连成一个网络，注:初始给定边无限

2>即选择n-1条边将n个结点连通

3>边无限即稠密图，进而选择最小生成树Prim算法(未堆优化)

3反思：memset()初始化注意第三个参数应为sizeof(初始化数组)

4思考：稠密图，Prim算法(未堆优化)

vjudge题目链接

以下为Accepted代码

1>16ms

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int inf = 0x3f3f3f3f;

int e[104][104], vis[104], dis[104];

void Prim(int n);

int main(){
int n, m, u, v, w;
while(scanf("%d", &n) && n){
scanf("%d", &m);

memset(e, inf, sizeof(e));

while(m--){
scanf("%d %d %d", &u, &v, &w);
if(e[u][v] > w)
e[u][v] = e[v][u] = w;
}

Prim(n);
}
return 0;
}
void Prim(int n){
int i, miv, v, num, sum;
memset(vis, 0, sizeof(vis));
for(i = 1; i <= n; i++)
dis[i] = e[1][i];
vis[1] = 1, dis[1] = 0, num = 1, sum = 0;
while(num < n){
miv = inf;
for(i = 1; i <= n; i++){
if(!vis[i] && dis[i] < miv){
miv = dis[i], v = i;
}
}
vis[v] = 1, num++, sum += dis[v];
for(i = 1; i <= n; i++){
if(!vis[i] && e[v][i] < dis[i])
dis[i] = e[v][i];
}
}
printf("%d\n", sum);
}