B

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);}