NYIST OJ 题目38 布线问题

最小生成树水题，先按最小生成树做，答案最后加上最小的从第i号楼接线到外界供电设施所需要的费用即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

const int maxn = 500;
struct abc{int uu, vv, cc;}node[maxn*(maxn - 1) / 2 + 10];
int uu[maxn*(maxn - 1) / 2 + 10];
int vv[maxn*(maxn - 1) / 2 + 10];
int cc[maxn*(maxn - 1) / 2 + 10];
int ff[maxn], father[maxn];
bool cmp(const abc&a, const abc&b) { return a.cc < b.cc; }

int find(int x)
{
    if (x != father[x]) father[x] = find(father[x]);
    return father[x];
}
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int i, n, m;
        scanf("%d%d", &n, &m);
        for (i = 0; i < m; i++) scanf("%d%d%d", &uu[i], &vv[i], &cc[i]);
        for (i = 0; i < n; i++) scanf("%d", &ff[i]);
        for (i = 0; i <= n; i++) father[i] = i;
        for (i = 0; i < m; i++)
        {
            node[i].uu = uu[i];
            node[i].vv = vv[i];
            node[i].cc = cc[i];
        }
        sort(node, node + m, cmp);
        sort(ff, ff + n);
        int anss = 0;
        for (i = 0; i < m; i++)
        {
            int fu = find(node[i].uu);
            int fv = find(node[i].vv);
            if (fu != fv)
            {
                father[fu] = fv;
                anss = anss + node[i].cc;
            }
        }
        printf("%d
", anss + ff[0]);
    }
    return 0;
}