第三章图论 No.4最小生成树的简单应用-Toy模板网

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存在边权为负的情况下，无法求最小生成树

第三章图论 No.4最小生成树的简单应用,AcWing算法提高课课程记录,图论

裸题：1140. 最短网络

1140. 最短网络 - AcWing题库
第三章图论 No.4最小生成树的简单应用,AcWing算法提高课课程记录,图论
套个prim的板子即可

#include <iostream>
#include <cstring>
using namespace std;

const int N = 110, INF = 0x3f3f3f3f;
int g[N][N];
int dis[N]; bool st[N];
int n;

int prim()
{
    memset(dis, 0x3f, sizeof(dis));
    int res = 0;
    for (int i = 0; i < n; ++ i )
    {
        int x = -1;
        for (int j = 1; j <= n; ++ j ) 
            if (!st[j] && (x == -1 || dis[x] > dis[j])) x = j;
        st[x] = true;
        if (i && dis[x] == INF) return INF;
        if (i) res += dis[x];
        
        for (int y = 1; y <= n; ++ y )
            dis[y] = min(dis[y], g[x][y]);
    }
    return res;
}

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; ++ i )
        for (int j = 1; j <= n; ++ j )
            scanf("%d", &g[i][j]);
            
    printf("%d\n", prim());
    return 0;
}

裸题：1141. 局域网

1141. 局域网 - AcWing题库
第三章图论 No.4最小生成树的简单应用,AcWing算法提高课课程记录,图论

裸题，稀疏图，套个kruskal的板子就行
需要注意的是：题目给定的图可能存在多个连通块，若使用prim算法，需要对每个连通块求最小生成树，但是使用kruskal能直接求出所有连通块的最小生成树

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 110, M = 210;
struct Edge
{
    int x, y, w;
    bool operator<(const Edge& e) const 
    {
        return w < e.w;
    }
}edges[M];

int p[N];
int n, m;

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

int kruskal()
{
    sort(edges, edges + m);
    for (int i = 1; i <= n; ++ i ) p[i] = i;
    int cnt = 0, res = 0;
    for (int i = 0; i < m; ++ i )
    {
        auto t = edges[i];
        int x = edges[i].x, y = edges[i].y, w = edges[i].w;
        x = find(x), y = find(y);
        if (x != y)
        {
            cnt ++ ;
            res += w;
            p[x] = y;
        }
    }
    return res;
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; ++ i )
        scanf("%d%d%d", &edges[i].x, &edges[i].y, &edges[i].w);
        
    int sum = 0;
    for (int i = 0; i < m; ++ i ) sum += edges[i].w;
    printf("%d\n", sum - kruskal());
    return 0;
}

裸题：1142. 繁忙的都市

1142. 繁忙的都市 - AcWing题库
依然是套kruskal的板子
第三章图论 No.4最小生成树的简单应用,AcWing算法提高课课程记录,图论

#include <iostream>
#include <algorithm>
using namespace std;

const int N = 310 ,M = 8010;
struct Edge
{
    int x, y, w;
    bool operator<(const Edge& e) const
    {
        return w < e.w;
    }
}edges[M];

int n, m;
int p[N];

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

int kruskal()
{
    sort(edges, edges + m);
    int res = 0;
    for (int i = 1; i <= n; ++ i ) p[i] = i;
    for (int i = 0; i < m; ++ i )
    {
        auto t = edges[i];
        int x = t.x, y = t.y, w = t.w;
        x = find(x), y = find(y);
        if (x != y)
        {
            res = max(res, w);
            p[x] = y;
        }
    }
    return res;
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; ++ i )
        scanf("%d%d%d", &edges[i].x, &edges[i].y, &edges[i].w);
        
    printf("%d %d\n", n - 1, kruskal());
    return 0;
}

裸题：1143. 联络员

1143. 联络员 - AcWing题库
第三章图论 No.4最小生成树的简单应用,AcWing算法提高课课程记录,图论

添加所有必选的边，维护并查集，然后再对非必选的边做kruskal


#include <iostream>
#include <algorithm>
using namespace std;

const int N = 2010, M = 10010;
struct Edge
{
    int x, y, w;
    bool operator<(const Edge& e) const 
    {
        return w < e.w;
    }
}edges[M];

int n, m, cnt;
int p[N];

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

int kruskal()
{
    int res = 0;
    for (int i = 0; i < cnt; ++ i )
    {
        auto t = edges[i];
        int x = t.x, y = t.y, w = t.w;
        x = find(x), y = find(y);
        if (x != y)
        {
            res += w;
            p[x] = y;
        }
    }
    return res;
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++ i ) p[i] = i;
    
    int t, x, y, d;
    int res = 0;
    while ( m -- )
    {
        scanf("%d%d%d%d", &t, &x, &y, &d);
        if (t == 1)
        {
            x = find(x), y = find(y);
            if (x != y) p[x] = y;
            res += d;
        }
        else
        {
            edges[cnt].x = x, edges[cnt].y = y, edges[cnt].w = d;
            cnt ++ ;
        }
    }
    sort(edges, edges + cnt);
    res += kruskal();
    printf("%d\n", res);
    
    return 0;
}

有些麻烦的裸题：1144. 连接格点

1144. 连接格点 - AcWing题库
第三章图论 No.4最小生成树的简单应用,AcWing算法提高课课程记录,图论

点阵为图中的点，将二维坐标转换成一维，作为点的编号
添加已有连线后，做kruskal即可

#include <iostream>
#include <algorithm>
using namespace std;

const int N = 1010;
struct Edge
{
    int x, y, w;
    bool operator<(const Edge& e) const 
    {
        return w < e.w;
    }
}edges[2 * N * N];

int n, m, cnt = 1;
int p[N * N];
int g[N][N]; // 二维到一维
int dx[3] = { 0, 1, 0 }, dy[3] = { 0, 0, 1 };

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

int kruskal()
{
    int res = 0;
    for (int i = 0; i < cnt; ++ i )
    {
        auto t = edges[i];
        int x = t.x, y = t.y, w = t.w;
        x = find(x), y = find(y);
        if (x != y)
        {
            res += w;
            p[x] = y;
        }
    }
    return res;
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++ i )
        for (int j = 1; j <= m; ++ j )
            g[i][j] = cnt ++ ;
    
    for (int i = 1; i < cnt; ++ i ) p[i] = i;
    int x1, x2, y1, y2;
    while (~scanf("%d%d%d%d", &x1, &y1, &x2, &y2))
    {
        int x = g[x1][y1], y = g[x2][y2];
        x = find(x), y = find(y);
        if (x != y) p[x] = y;
    }
    
    cnt = 0;
    for (int i = 1; i <= n; ++ i )
        for (int j = 1; j <= m; ++ j )
            for (int k = 1; k <= 2; ++ k )
            {
                int a = i + dx[k], b = j + dy[k];
                if (a >= 1 && a <= n && b >= 1 && b <= m)
                {
                    int x = g[i][j], y = g[a][b];
                    edges[cnt ++ ] = { x, y, k };
                }
                    
            }
            
    sort(edges, edges + cnt);
    printf("%d\n", kruskal());
    
    return 0;
}

debug：n * m的矩阵中，相邻两点之间存在一条边，那么矩阵中的边数应该为m(n-1) + n(m-1)，大概就是2 * n * n，数组开小了导致SF
尽量不要在for循环中定义除了循环变量之外的变量

第三章图论 No.4最小生成树的简单应用,AcWing算法提高课课程记录,图论
需要注意的是，200万条边进行排序会消耗很多时间，由于边的权值只有1和2，所以可以先添加权值为1的边，再添加权值为2的边文章来源地址https://www.toymoban.com/news/detail-628344.html