acwing算法提高之图论--最小生成树的典型应用

1 介绍

本专题用来记录使用prim算法或kruskal算法求解的题目。

2 训练

题目1：1140最短网络

C++代码如下，

#include <iostream>
#include <cstring>

using namespace std;

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

void prim() {
    memset(d, 0x3f, sizeof d);
    
    int res = 0;
    for (int i = 0; i < n; ++i) {
        int t = -1;
        for (int j = 1; j <= n; ++j) {
            if (!st[j] && (t == -1 || d[t] > d[j])) {
                t = j;
            }
        }
        
        st[t] = true;
        if (i) res += d[t];
        
        for (int j = 1; j <= n; ++j) {
            if (d[j] > g[t][j]) {
                d[j] = g[t][j];
            }
        }
    }
    
    cout << res << endl;
    return;
}

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

题目2：1141局域网

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 110, M = 210;
int p[N];
int n, m;

struct Edge {
    int a, b, w;
    bool operator< (const Edge &W) const {
        return w < W.w;
    }
}edges[M];

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

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

题目3：1142繁忙的都市

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 310, M = 8010;
int n, m;
int p[N];

struct Edge {
    int a, b, w;
    bool operator< (const Edge &W) const {
        return w < W.w;
    }
}edges[M];

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

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

题目4：