先用kruskal演算法算出最小生成樹，並把最小生成樹的邊記錄下來。然後依次列舉刪除邊，用其他的邊再次使用kruskal演算法算出最小生成樹。如果算出的代價和原來的相同，則不唯一，否則唯一。另外當我們刪除一條邊之後，可能根本構不成一顆生成樹，要判斷一下。

程式碼如下：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <queue>
#include <new>
#include <set>
#include <map>

using namespace std;

const int maxn = 1000;
int p[maxn], x[maxn], y[maxn], n;
struct edge
{
    int u, v, cost;
    bool operator < (const edge& b) const
    {
        return cost < b.cost;
    }
};
vector<edge> edges;
vector<int> mst_edges;

int find_root(int x)
{
    if (p[x] == -1) return x;
    return p[x] = find_root(p[x]);
}

void kruskal()
{
    memset(p, -1, sizeof(p));
    sort(edges.begin(), edges.end());

    mst_edges.clear();
    int cost = 0;
    for (int i = 0; i < edges.size(); i++)
    {
        int x = edges[i].u, y = edges[i].v;
        int rx = find_root(x);
        int ry = find_root(y);
        if (rx != ry)
        {
            p[rx] = ry;
            cost += edges[i].cost;
            mst_edges.push_back(i);
        }
    }

    printf("%d\n", cost);

    bool ok = false;
    for (int i = 0; i < mst_edges.size(); i++)
    {
        memset(p, -1, sizeof(p));
        int ans = 0, k = 0;
        for (int j = 0; j < edges.size(); j++)
        {
            if (mst_edges[i] == j) continue;
            int x = edges[j].u, y = edges[j].v;
            int rx = find_root(x);
            int ry = find_root(y);
            if (rx != ry)
            {
                p[rx] = ry;
                ans += edges[j].cost;
                k++;
            }
        }

        if (ans == cost && k == n - 1)//新算出的生成樹的代價和最小生成樹代價相同，且能構成一棵樹（最小生成樹的邊數肯定是n - 1）
        {
            ok = true;
            break;
        }
    }

    if (ok) printf("Yes\n");
    else printf("No\n");
}

int main()
{
    //freopen("1.txt", "r", stdin);
    int T, Case = 0;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        edges.clear();
        for (int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
        for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
        {
            int d = abs(x[i] - x[j]) + abs(y[i] - y[j]);
            edges.push_back(edge{i, j, d});
        }

        printf("case #%d:\n", Case++);

        kruskal();
    }
    return 0;
}