DS|图（连通与生成树）

题目一：DS图 -- 图的连通分量

题目描述：

输入无向图顶点信息和边信息，创建图的邻接矩阵存储结构，计算图的连通分量个数。

输入要求：

测试次数t

每组测试数据格式如下：

第一行：顶点数 顶点信息

第二行：边数

第三行开始，每行一条边信息

输出要求：

每组测试数据输出，顶点信息和邻接矩阵信息

输出图的连通分量个数，具体输出格式见样例。

每组输出直接用空行分隔。

输入样例：

3
4 A B C D
2
A B
A C
6 V1 V2 V3 V4 V5 V6
5
V1 V2
V1 V3
V2 V4
V5 V6
V3 V5
8 1 2 3 4 5 6 7 8
5
1 2
1 3
5 6
5 7
4 8

输出样例：

A B C D
0 1 1 0
1 0 0 0
1 0 0 0
0 0 0 0
2

V1 V2 V3 V4 V5 V6
0 1 1 0 0 0
1 0 0 1 0 0
1 0 0 0 1 0
0 1 0 0 0 0
0 0 1 0 0 1
0 0 0 0 1 0
1

1 2 3 4 5 6 7 8
0 1 1 0 0 0 0 0
1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1
0 0 0 0 0 1 1 0
0 0 0 0 1 0 0 0
0 0 0 0 1 0 0 0
0 0 0 1 0 0 0 0
3

代码示例：

#include <iostream>
#include <string>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;

class Map {
private:
    int** array;
    string* vertex;
    int n;
    bool visited[20];
    int partnumber;
public:
    Map() {
        cin >> n;
        array = new int* [n];
        for (int i = 0; i < n; i++) {
            array[i] = new int[n];
            for (int j = 0; j < n; j++) array[i][j] = 0;
        }
        vertex = new string[n];
        for (int i = 0; i < n; i++) cin >> vertex[i];
        partnumber = 0;
    }

    int findIndex(string str) {
        for (int i = 0; i < n; i++) if (str == vertex[i]) return i;
        return -1;
    }

    void createMap() {
        int n, v1, v2;
        string ch1, ch2;
        cin >> n;
        for (int i = 0; i < n; i++) {
            cin >> ch1 >> ch2;
            v1 = findIndex(ch1);
            v2 = findIndex(ch2);
            array[v1][v2] = 1, array[v2][v1] = 1;
        }
    }
    void DFSTraverse() {
        for (int i = 0; i < n; i++) visited[i] = false;
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                DFS(i);
                partnumber++;
            }
        }
    }
    void DFS(int v) {
        visited[v] = true;
        int* temp = new int[n];
        int pos = 0;
        for (int i = 0; i < n; i++) if (array[v][i]) temp[pos++] = i;
        for (int i = 0; i < pos; i++) if (!visited[temp[i]]) DFS(temp[i]);
    }
    void printMap() {
        for (int i = 0; i < n; i++) {
            cout << vertex[i];
            if (i == n - 1) cout << endl;
            else cout << " ";
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cout << array[i][j];
                if (j == n - 1) cout << endl;
                else cout << " ";
            }
        }
        cout << partnumber << endl;
    }
    ~Map() {
        for (int i = 0; i < n; i++) delete[]array[i];
        delete[]array;
        delete[]vertex;
    }
};

int main() {
    int t;
    cin >> t;
    while (t--) {
        Map M;
        M.createMap();
        M.DFSTraverse();
        M.printMap();
        cout << endl;
    }
    return 0;
}

题目二：DS图 -- 最小生成树

题目描述：

根据输入创建无向网。分别用Prim算法和Kruskal算法构建最小生成树。（假设：输入数据的最小生成树唯一。）

输入要求：

顶点数n

n个顶点

边数m

m条边信息,格式为：顶点1顶点2权值

Prim算法的起点v

输出要求：

输出最小生成树的权值之和

对两种算法，按树的生长顺序，输出边信息(Kruskal中边顶点按数组序号升序输出)

输入样例：

6
v1 v2 v3 v4 v5 v6 
10
v1 v2 6
v1 v3 1
v1 v4 5
v2 v3 5
v2 v5 3
v3 v4 5
v3 v5 6
v3 v6 4
v4 v6 2
v5 v6 6
v1

输出样例：

15
prim:
v1 v3 1
v3 v6 4
v6 v4 2
v3 v2 5
v2 v5 3
kruskal:
v1 v3 1
v4 v6 2
v2 v5 3
v3 v6 4
v2 v3 5

代码示例：

#include <iostream>
#include <string>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;

const int MAXNUMBER = 100;

struct Close {
    int adjvex;
    int loweight = 0x3f3f3f3f;
} closedge[MAXNUMBER];

struct Path {
    int head;
    int tail;
};

class Map {
    int matrix[MAXNUMBER][MAXNUMBER] = { 0 };
    string* vertex;
    int n;//顶点数
    int leastWeight = 0;
    vector<Path> path;
    int shortdis[MAXNUMBER];
public:
    Map() {
        cin >> n;
        vertex = new string[n];
        for (int i = 0; i < n; i++) cin >> vertex[i];
        for (int i = 0; i < n; i++) shortdis[i] = i;
        int edge;
        cin >> edge;
        for (int i = 0; i < edge; i++) {
            string vex1, vex2;
            int weight;
            cin >> vex1 >> vex2 >> weight;
            matrix[find(vex1)][find(vex2)] = matrix[find(vex2)][find(vex1)] = weight;
        }
    }

    int find(string str) {
        for (int i = 0; i < n; i++)
            if (vertex[i] == str) return i;
    }

    void Prim() {
        string start_vertex;
        cin >> start_vertex;
        int start = find(start_vertex);

        for (int i = 0; i < n; i++) {
            if (matrix[start][i]) closedge[i] = { start, matrix[start][i] };
            else closedge[i].adjvex = start;
        }

        closedge[start].loweight = 0;
        for (int i = 1; i < n; i++) {
            int minWeight = 0x3f3f3f3f, nextedge;
            for (int j = 0; j < n; j++)
                if (minWeight > closedge[j].loweight && closedge[j].loweight) {
                    minWeight = closedge[j].loweight;
                    nextedge = j;
                }
            closedge[nextedge].loweight = 0;
            Path p = { closedge[nextedge].adjvex, nextedge };
            path.push_back(p);
            leastWeight += minWeight;
            for (int j = 0; j < n; j++)
                if (closedge[j].loweight > matrix[nextedge][j] && matrix[nextedge][j])
                    closedge[j] = { nextedge, matrix[nextedge][j] };
        }
        cout << leastWeight << endl;
        cout << "prim:" << endl;
        for (auto& item : path)
            cout << vertex[item.head] << ' ' << vertex[item.tail] << ' ' << matrix[item.head][item.tail] << endl;
    }
    int get(int x) {
        if (shortdis[x] == x) return x;
        return get(shortdis[x]);
    }
    void Kruskal() {
        path.clear();
        cout << "kruskal:" << endl;
        int cnt = 0;
        for (int k = 0; k < n - 1; k++) {
            int minWeight = 0x3f3f3f3f, tail = 0, head = 0;
            for (int i = 0; i < n; i++) {
                for (int j = i + 1; j < n; j++) {
                    if (matrix[i][j] && minWeight > matrix[i][j] && get(i) != get(j)) {
                        head = i, tail = j;
                        minWeight = matrix[i][j];
                    }
                }
            }

            shortdis[get(tail)] = get(head);
            Path p = { head, tail };
            path.push_back(p);
            leastWeight += minWeight;
            if (tail || head) cnt++;
        }
        if (cnt == n - 1) {
            //cout << leastWeight << endl;
            for (auto& item : path) cout << vertex[item.head] << ' ' << vertex[item.tail] << ' ' << matrix[item.head][item.tail] << endl;
        }
        else cout << "-1" << endl;
    }
};

int main() {
    Map map;
    map.Prim();
    map.Kruskal();
}

题目三：DS图 -- 汉密尔顿回路

题目描述：

著名的“汉密尔顿（Hamilton）回路问题”是要找一个能遍历图中所有顶点的简单回路（即每个顶点只访问 1 次）。本题就要求你判断任一给定的回路是否汉密尔顿回路。

输入要求：

首先第一行给出两个正整数：无向图中顶点数?N（2<N≤200）和边数?M。随后?M?行，每行给出一条边的两个端点，格式为“顶点1 顶点2”，其中顶点从 1 到N?编号。再下一行给出一个正整数?K，是待检验的回路的条数。随后?K?行，每行给出一条待检回路，格式为：

n?V1??V2????Vn?

其中?n?是回路中的顶点数，Vi??是路径上的顶点编号。

输出要求：

对每条待检回路，如果是汉密尔顿回路，就在一行中输出"YES"，否则输出"NO"。

输入样例：

6 10
6 2
3 4
1 5
2 5
3 1
4 1
1 6
6 3
1 2
4 5
6
7 5 1 4 3 6 2 5
6 5 1 4 3 6 2
9 6 2 1 6 3 4 5 2 6
4 1 2 5 1
7 6 1 3 4 5 2 6
7 6 1 2 5 4 3 1

输出样例：

YES
NO
NO
NO
YES
NO

代码示例：

#include <iostream>
#include <string>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;

class Map {
private:
    int** array;
    int* vertex;
    int n;
    bool visited[20];
    int partnumber;
public:
    Map() {
        cin >> n;
        array = new int* [n];
        for (int i = 0; i < n; i++) {
            array[i] = new int[n];
            for (int j = 0; j < n; j++) array[i][j] = 0;
        }
        vertex = new int[n];
        for (int i = 0; i < n; i++) vertex[i] = i + 1;
        partnumber = 0;
    }

    int findIndex(int num) {
        for (int i = 0; i < n; i++) if (num == vertex[i]) return i;
        return -1;
    }

    void createMap() {
        int m, v1, v2;
        int num1, num2;
        cin >> m;
        for (int i = 0; i < m; i++) {
            cin >> num1 >> num2;
            v1 = findIndex(num1);
            v2 = findIndex(num2);
            array[v1][v2] = 1, array[v2][v1] = 1;
        }
    }
    void DFSTraverse() {
        for (int i = 0; i < n; i++) visited[i] = false;
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                DFS(i);
                partnumber++;
            }
        }
    }
    void DFS(int v) {
        visited[v] = true;
        int* temp = new int[n];
        int pos = 0;
        for (int i = 0; i < n; i++) if (array[v][i]) temp[pos++] = i;
        for (int i = 0; i < pos; i++) if (!visited[temp[i]]) DFS(temp[i]);
    }
    void printMap() {
        for (int i = 0; i < n; i++) {
            cout << vertex[i];
            if (i == n - 1) cout << endl;
            else cout << " ";
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cout << array[i][j];
                if (j == n - 1) cout << endl;
                else cout << " ";
            }
        }
        cout << partnumber << endl;
    }
    void Hamilton() {
        int k;
        bool mark = true;
        int cnt[20] = { 0 };
        cin >> k;
        int head, tail;
        cin >> head;
        int start = head;
        for (int i = 1; i < k; i++) {
            cin >> tail;
            cnt[tail]++;
            if (i != k - 1 && tail == start || cnt[tail] == 2 && tail != start) mark = false;
            if (!array[findIndex(head)][findIndex(tail)]){
                cout << "NO" << endl;
                continue;
            }
            head = tail;
            if (start == tail && i == k - 1 && k >= n && mark) cout << "YES" << endl;
            else if (start != tail && i == k - 1 || k < n && i == k - 1 || !mark && i == k - 1) cout << "NO" << endl;
        }
        
    }
    ~Map() {
        for (int i = 0; i < n; i++) delete[]array[i];
        delete[]array;
        delete[]vertex;
    }
};

int main() {
    Map M;
    M.createMap();
    int t;
    cin >> t;
    while (t--) M.Hamilton();
    return 0;
}

题目四：DS图 -- 六度空间

题目描述：

“六度空间”理论又称作“六度分隔（Six Degrees of Separation）”理论。这个理论可以通俗地阐述为：“你和任何一个陌生人之间所间隔的人不会超过六个，也就是说，最多通过五个人你就能够认识任何一个陌生人。”

“六度空间”理论虽然得到广泛的认同，并且正在得到越来越多的应用。但是数十年来，试图验证这个理论始终是许多社会学家努力追求的目标。然而由于历史的原因，这样的研究具有太大的局限性和困难。随着当代人的联络主要依赖于电话、短信、微信以及因特网上即时通信等工具，能够体现社交网络关系的一手数据已经逐渐使得“六度空间”理论的验证成为可能。

假如给你一个社交网络图，请你对每个节点计算符合“六度空间”理论的结点占结点总数的百分比。

输入要求：

输入第1行给出两个正整数，分别表示社交网络图的结点数N（??，表示人数）、边数M（≤33×N，表示社交关系数）。随后的M行对应M条边，每行给出一对正整数，分别是该条边直接连通的两个结点的编号（节点从1到N编号）。

输出要求：

对每个结点输出与该结点距离不超过6的结点数占结点总数的百分比，精确到小数点后2位。每个结节点输出一行，格式为“结点编号:（空格）百分比%”。

输入样例：

10 9
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10

输出样例：

1: 70.00%
2: 80.00%
3: 90.00%
4: 100.00%
5: 100.00%
6: 100.00%
7: 100.00%
8: 90.00%
9: 80.00%
10: 70.00%

代码示例：

#include <iostream>
#include <string>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;

class Map {
private:
    int** array;
    int* vertex;
    int n;
    bool visited[20];
    int partnumber;
    int** reach;
public:
    Map() {
        cin >> n;
        array = new int* [n];
        reach = new int* [n];
        for (int i = 0; i < n; i++) {
            array[i] = new int[n];
            reach[i] = new int[n];
            for (int j = 0; j < n; j++) array[i][j] = 0, reach[i][j] = 0;
        }
        vertex = new int[n];
        for (int i = 0; i < n; i++) vertex[i] = i + 1;
        partnumber = 0;
    }

    int findIndex(int num) {
        for (int i = 0; i < n; i++) if (num == vertex[i]) return i;
        return -1;
    }

    void createMap() {
        int m, v1, v2;
        int num1, num2;
        cin >> m;
        for (int i = 0; i < m; i++) {
            cin >> num1 >> num2;
            v1 = findIndex(num1);
            v2 = findIndex(num2);
            array[v1][v2] = 1, array[v2][v1] = 1;
        }
    }

    void SDS() {
        for (int i = 0; i < n; i++) {
            int number = SDSBFS(i);
            cout << i + 1 << ": " << fixed << setprecision(2) << (double)number * 100 / n << "%" << endl;
            memset(visited, false, sizeof(visited));
        }
    }
    int SDSBFS(int i) {
        int cnt = 0;
        int level = 0, last = i, tail = -1;
        queue<int> q;
        visited[i] = true;
        q.push(i);
        cnt++;
        while (!q.empty()) {
            int k = q.front();
            q.pop();
            for (int j = 0; j < n; j++)
            {
                if (j == k) continue;
                if (!visited[j] && array[k][j]) {
                    q.push(j);
                    tail = j;
                    visited[j] = true;
                    cnt++;
                }
            }
            if (k == last) level++, last = tail;
            if (level == 6) break;
        }
        return cnt;
    }
    ~Map() {
        for (int i = 0; i < n; i++) delete[]array[i];
        delete[]array;
        delete[]vertex;
    }
};

int main() {
    Map M;
    M.createMap();
    M.SDS();
}

题目五：DS图 --?红色警报

题目描述：

战争中保持各个城市间的连通性非常重要。本题要求你编写一个报警程序，当失去一个城市导致国家被分裂为多个无法连通的区域时，就发出红色警报。注意：若该国本来就不完全连通，是分裂的k个区域，而失去一个城市并不改变其他城市之间的连通性，则不要发出警报。

输入要求：

输入在第一行给出两个整数N（0?<?N?≤?500）和M（≤?5000），分别为城市个数（于是默认城市从0到N-1编号）和连接两城市的通路条数。随后M行，每行给出一条通路所连接的两个城市的编号，其间以1个空格分隔。在城市信息之后给出被攻占的信息，即一个正整数K和随后的K个被攻占的城市的编号。

注意：输入保证给出的被攻占的城市编号都是合法的且无重复，但并不保证给出的通路没有重复。

输出要求：

对每个被攻占的城市，如果它会改变整个国家的连通性，则输出Red Alert: City k is lost!，其中k是该城市的编号；否则只输出City k is lost.即可。如果该国失去了最后一个城市，则增加一行输出Game Over.。

输入样例：

5 4
0 1
1 3
3 0
0 4
5
1 2 0 4 3

输出样例：

City 1 is lost.
City 2 is lost.
Red Alert: City 0 is lost!
City 4 is lost.
City 3 is lost.
Game Over.

代码示例：

#include <iostream>
#include <string>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;

class Map {
private:
    int** array;
    int* vertex;
    int n;
    bool visited[20];
    int partnumber;
    int** reach;
    int origin[20];
public:
    Map() {
        cin >> n;
        array = new int* [n];
        reach = new int* [n];
        for (int i = 0; i < n; i++) {
            array[i] = new int[n];
            reach[i] = new int[n];
            for (int j = 0; j < n; j++) array[i][j] = 0, reach[i][j] = 0;
        }
        vertex = new int[n];
        for (int i = 0; i < n; i++) vertex[i] = i;
        partnumber = 0;
    }

    int findIndex(int num) {
        for (int i = 0; i < n; i++) if (num == vertex[i]) return i;
        return -1;
    }

    void createMap() {
        int m, v1, v2;
        int num1, num2;
        cin >> m;
        for (int i = 0; i < m; i++) {
            cin >> num1 >> num2;
            v1 = findIndex(num1);
            v2 = findIndex(num2);
            array[v1][v2] = 1, array[v2][v1] = 1;
        }
    }

    void RA() {
        memset(origin, 0, sizeof(visited));

        for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if (array[i][j]) origin[i]++;

        int x;
        cin >> x;
        while (x--) {
            int y;
            cin >> y;
            int tozerocnt = 0;
            for (int i = 0; i < n; i++) if (array[y][i]) array[y][i] = 0, array[i][y] = 0;
            int cur[20] = { 0 };
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) if (array[i][j]) cur[i]++;
            }
            for (int i = 0; i < n; i++) {
                if (origin[i] != 0 && cur[i] == 0) tozerocnt++;
                origin[i] = cur[i];
            }
            if (tozerocnt > 1) cout << "Red Alert: City " << y << " is lost!" << endl;
            else cout << "City " << y << " is lost." << endl;
        }
        cout << "Game Over." << endl;
    }
    ~Map() {
        for (int i = 0; i < n; i++) delete[]array[i];
        delete[]array;
        delete[]vertex;
    }
};

int main() {
    Map M;
    M.createMap();
    M.RA();
}