寒假：Day23

寒假：Day23 Day23

最小生成树开搞！

1140. 最短网络 - AcWing题库

p r i m prim prim 模板题，因为数据量小，而且是以邻接矩阵的方式给出图，所以用 p r i m prim prim 算法更加方便

#include
using namespace std;
const int N = 110;
int w[N][N];
int dist[N];
bool st[N];
int n;

int prim()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    int res = 0;
    for (int i = 0; i < n; i++)
    {
        int t = -1;
        for (int j = 1; j <= n; j++) // 找到距离生成树最近的点
            if (!st[j] && (t == -1 || dist[t] > dist[j])) t = j;
            
        res += dist[t]; // 把这个点加入生成树
        st[t] = true;
        
        for (int j = 1; j <= n; j++) // 用这个点更新其他点
            dist[j] = min(dist[j], w[t][j]);
    }
    
    return res;
}

int main(void)
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            cin >> w[i][j];
    cout << prim() << endl ;
    return 0;
}

1141. 局域网 - AcWing题库

K r u s k a l Kruskal Kruskal 模板题，如果已经在一个集合的边就删掉

#include
using namespace std;
const int N = 110, M = 210;

int n, m;
struct Edge
{
    int a, b, w;
    bool operator< (const Edge &t) const // 重载运算符
    {
        return w < t.w;
    }
}e[M];
int p[N];

int find(int x) // 并查集模板
{
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main(void)
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++) p[i] = i;
    for (int i = 0; i < m; i++) 
    {
        int a, b, w;
        cin >> a >> b >> w;
        e[i] = {a, b, w};
    }
    sort(e, e + m); // 把边从小到大排序
    int res = 0;
    for (int i = 0; i < m; i++)
    {
        int a = find(e[i].a), b = find(e[i].b), w = e[i].w;
        if (a != b) p[a] = b; // 如果他们两个不在一个集合中，那就加到一起
        else res += w; // 如果是，那就把这条边删掉
    }
    cout << res << endl ;
    return 0;
}

1142. 繁忙的都市 - AcWing题库

K r u s k a l Kruskal Kruskal 模板题，找最小生成树的最长边即可

#include
using namespace std;
const int N = 310, M = 8010;

int n, m;
struct Edge
{
    int a, b, w;
    bool operator< (const Edge &t) const
    {
        return w < t.w;
    }
}e[M];
int p[N];

int find(int x)
{
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main(void)
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++) p[i] = i;
    for (int i = 0; i < m; i++) 
    {
        int a, b, w;
        cin >> a >> b >> w;
        e[i] = {a, b, w};
    }
    sort(e, e + m); 
    int res = 0;
    for (int i = 0; i < m; i++)
    {
        int a = find(e[i].a), b = find(e[i].b), w = e[i].w;
        if (a != b) {
            p[a] = b; 
            res = max(res, w);
        }
    }
    cout << n - 1 << " " << res << endl ;
    return 0;
}

1143. 联络员 - AcWing题库

还是 K r u s k a l Kruskal Kruskal 模板题，根据题意套模板就行

#include
using namespace std;
const int N = 2010, M = 10010;

int n, m, cnt, res;
int P[N];
struct Edge
{
    int a, b, w;
    bool operator< (const Edge &t) const
    {
        return w < t.w;
    }
}e[M];

int find(int x)
{
    if (x != P[x]) P[x] = find(P[x]);
    return P[x];
}

int main(void)
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++) P[i] = i;
    for (int i = 0; i < m; i++)
    {
        int p, a, b, w;
        cin >> p >> a >> b >> w;
        if (p == 1) {
            res += w;
            a = find(a), b = find(b);
            if (a != b) P[a] = b;
        }
        else {
            e[cnt++] = {a, b, w};
        } 
    }
    sort(e, e + cnt);
    for (int i = 0; i < cnt; i++)
    {
        int a = find(e[i].a), b = find(e[i].b), w = e[i].w;
        if (a != b)
        {
            P[a] = b;
            res += w;
        }
    }
    cout << res << endl ;
    return 0;
}

1146. 新的开始 - AcWing题库

这题给了邻接矩阵，应该是用 p r i m prim prim 算法，但是我顺手写的是 K r u s k a l Kruskal Kruskal 算法，做法也很简单，设一个超级源点，然后让这个源点向每一个发电机连建边，然后跑一边模板即可。

#include
using namespace std;
const int N = 310, M = 100010;

int n, p[N], cnt, x;
struct Edge
{
    int a, b, w;
    bool operator< (const Edge &t) const
    {
        return w < t.w;
    }
}e[M + N];

int find(int x)
{
    if (x != p[x]) p[x] = find(p[x]);
    return p[x];
}

int main(void)
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 0; i <= n; i++) p[i] = i;
    for (int i = 1; i <= n; i++) {
        cin >> x;
        e[cnt++] = {0, i, x};
    }
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
        {
            cin >> x;
            e[cnt++] = {i, j, x};
        }
    sort(e, e + cnt);
    int res = 0;
    for (int i = 0; i < cnt; i++)
    {
        int a = find(e[i].a), b = find(e[i].b), w = e[i].w;
        if (a != b)
        {
            res += w;
            p[a] = b;
        }
    }
    cout << res << endl ;
    return 0;
}

1145. 北极通讯网络 - AcWing题库

现在给我弄得就只会写 K r u s k a l Kruskal Kruskal 算法了， K r u s k a l Kruskal Kruskal 真好用，每一次连边都是集合的合并，所以当集合不超过 k k k 个时就输出当前边即可

#include
#define x first
#define y second
using namespace std;
const int N = 510, M = N * N;
typedef pair PII;

PII v[N];
int n, m, p[N], cnt;
struct Edge
{
    int a, b;
    double w;
    bool operator< (const Edge &t) const
    {
        return w < t.w;
    }
}e[M];

double get(PII a, PII b)
{
    int dx = a.x - b.x, dy = a.y - b.y;
    return sqrt(dx * dx + dy * dy);
}

int find(int x)
{
    if (x != p[x]) p[x] = find(p[x]);
    return p[x];
}

int main(void)
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++) p[i] = i;
    for (int i = 1; i <= n; i++) cin >> v[i].x >> v[i].y;
    for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
            e[cnt++] = {i, j, get(v[i], v[j])};

    sort(e, e + cnt);
    
    for (int i = 0; i < cnt; i++) 
    {
        int a = find(e[i].a), b = find(e[i].b);
        double w = e[i].w;
        if (a != b)
        {
            p[a] = b;
            if (-- n == m) {
                printf("%.2f", w);
            }
        }
    }
    return 0;
}