acwing算法提高之图论--最小生成树的扩展应用

1 介绍

本专题用来记录使用最小生成树算法（prim或kruskal）解决的扩展题目。

2 训练

C++代码如下，

cpp 复制代码

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

using namespace std;

const int N = 310, INF = 0x3f3f3f3f3;
int n, m;
int g[N][N];
int d[N];
bool st[N];

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

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

    return 0;   
}

题目2 ：1145北极通讯网络

C++代码如下，

cpp 复制代码

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;

typedef pair<int, int> PII;

const int N = 510, M = N * N;
int n, k;
vector<PII> points;
int p[N];

struct Edge {
    int a, b;
    double 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];
}

double compute_dis(int i, int j) {
    int x1 = points[i].first, y1 = points[i].second;
    int x2 = points[j].first, y2 = points[j].second;
    
    double d = double(x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
    d = sqrt(d);
    return d;
}

int main() {
    cin >> n >> k;
    for (int i = 0; i < n; ++i) {
        int x, y;
        cin >> x >> y;
        points.emplace_back(x,y);
    }
    
    for (int i = 0; i < n; ++i) p[i] = i;
    
    int m = 0;
    for (int i = 0; i < n; ++i) {
        for (int j = i + 1; j < n; ++j) {
            double w = compute_dis(i, j);
            edges[m] = {i, j, w};
            m += 1;
        }
    }
    
    sort(edges, edges + m);
    
    double res = 0.0;
    int cnt = n; //连通块的个数
    for (int i = 0; i < m; ++i) {
        int a = edges[i].a, b = edges[i].b;
        double w = edges[i].w;
        a = find(a);
        b = find(b);
        
        if (cnt == k) {
            break;
        }
        
        if (a != b) {
            p[a] = b;
            cnt--;
            res = w;
        }
    }
    
    printf("%.2lf\n", res);
    
    return 0;
}

题目3 ：346走廊泼水节

C++代码如下，

cpp 复制代码

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

using namespace std;

const int N = 6010;
int n, m;
int p[N], ms[N];
struct Edge {
    int a, b, w;
    bool operator< (const Edge &W) const {
        return w < W.w;
    }
}edges[N];

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

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

题目4 ：1148秘密的牛奶运输

C++代码如下，

cpp 复制代码

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

using namespace std;

typedef long long LL;

const int N = 510, M = 10010;
int n, m;
struct Edge {
    int a, b, w;
    bool f;
    bool operator< (const Edge &W) const {
        return w < W.w;
    }
}edges[M];
int p[N];
int dist1[N][N], dist2[N][N];
int h[N], e[N * 2], w[N * 2], ne[N * 2], idx;

void add(int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

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

void dfs(int u, int fa, int maxd1, int maxd2, int d1[], int d2[]) {
    d1[u] = maxd1, d2[u] = maxd2;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j != fa) {
            int td1 = maxd1, td2 = maxd2;
            if (w[i] > td1) td2 = td1, td1 = w[i];
            else if (w[i] < td1 && w[i] > td2) td2 = w[i];
            dfs(j, u, td1, td2, d1, d2);
        }
    }
    return;
}

int main() {
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof h);
    for (int i = 0; i < m; ++i) {
        int a, b, w;
        scanf("%d%d%d", &a, &b, &w);
        edges[i] = {a, b, w};
    }
    
    sort(edges, edges + m);
    for (int i = 1; i <= n; ++i) p[i] = i;
    
    LL sum = 0;
    for (int i = 0; i < m; ++i) {
        int a = edges[i].a, b = edges[i].b, w = edges[i].w;
        int pa = find(a), pb = find(b);
        if (pa != pb) {
            p[pa] = pb;
            sum += w;
            add(a, b, w), add(b, a, w);
            edges[i].f = true;
        }
    }
    
    for (int i = 1; i <= n; ++i) dfs(i, -1, -1e9, -1e9, dist1[i], dist2[i]);
    
    LL res = 1e18;
    for (int i = 0; i < m; ++i) {
        if (!edges[i].f) {
            int a = edges[i].a, b = edges[i].b, w = edges[i].w;
            LL t;
            if (w > dist1[a][b]) {
                t = sum + w - dist1[a][b];
            } else if (w > dist2[a][b]) {
                t = sum + w - dist2[a][b];
            }
            res=  min(res, t);
        }
    }
    
    printf("%lld\n", res);
    
    return 0;
}

acwing算法提高之图论--最小生成树的扩展应用

目录

1 介绍

2 训练