DS|图（连通与生成树）-编程知识

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
2V1 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
11 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（ $1\leq N\leq 10^3$ ，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：

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：

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();
}