蓝桥杯 城市建设

思路：

将河看做一个额外的点，将修码头看成是在相应地点和河之间连边，然后考虑是否修码头两种情况用最小生成树求解。需要注意的是如果边的权值是负数，那么即是已经连通，也要加这条边。

实现：

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 10005;
const int MAXM = 100005;
const int INF = 0x3f3f3f3f;
struct node
{
    int x, y, c;
};
node a[MAXN + MAXM];
int par[MAXN];
void init(int n)
{
    for (int i = 1; i <= n; i++) par[i] = i;
}
int find(int x)
{
    if (x == par[x]) return x;
    return par[x] = find(par[x]);
}
bool same(int x, int y)
{
    return find(x) == find(y);
}
void uni(int x, int y)
{
    x = find(x); y = find(y);
    if (x != y) par[x] = y;
}
bool cmp(node & a, node & b)
{
    return a.c < b.c;
}
int kru(int n, int m)
{
    init(n);
    sort(a + 1, a + m + 1, cmp);
    int cost = 0, cnt = n;
    for (int i = 1; i <= m; i++)
    {
        if (!same(a[i].x, a[i].y))
        {
            uni(a[i].x, a[i].y);
            cnt--;
            cost += a[i].c;
        }
        else if (a[i].c < 0) cost += a[i].c;
    }
    return cnt == 1 ? cost : INF;
}
int main()
{
    int n, m, w;
    cin >> n >> m;
    for (int i = 1; i <= m; i++) cin >> a[i].x >> a[i].y >> a[i].c;
    int cost = kru(n, m);
    int now = m + 1;
    for (int i = 1; i <= n; i++)
    {
        cin >> w;
        if (w == -1) continue;
        a[now].x = i; a[now].y = n + 1; a[now++].c = w;
    }
    int cost1 = kru(n + 1, now - 1);
    cout << min(cost, cost1) << endl;
    return 0;
}