题目如下：
It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of the cities
connected. Given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be repaired, quickly.
For example, if we have 3 cities and 2 highways connecting city1-city2 and city1-city3. Then if city1 is
occupied by the enemy, we must have 1 highway repaired, that is the highway city2-city3.
Input
Each input file contains one test case. Each case starts with a line containing 3 numbers N (<1000), M and K, which are the total number of cities, the number of remaining highways, and the number of cities to be checked, respectively. Then M lines follow,
each describes a highway by 2 integers, which are the numbers of the cities the highway connects. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.
Output
For each of the K cities, output in a line the number of highways need to be repaired if that city is lost.
Sample Input
3 2 3
1 2
1 3
1 2 3
Sample Output
1
0
0
看到这个题目，我的第一想法是用DFS来处理，利用DFS以每个结点为起点进行搜索，找出不能访问到所有结点的结点，这些结点构成的独立区域便是那些需要额外修建公路的区域，但有个问题是如果独立区域有多个结点，在计数时会出现重复，但是可以发现，在对某个结点进行DFS时，所有和它连通的结点都被访问到了，因此不必担心重复计数的问题。最后独立区域的个数-1即为要修建公路的条数。
#include<stdio.h>
#include<string.h> #define max 1001
int edge[max][max];
int visited[max];
int N, M, K; void DFS(int start)
{
visited[start] = 1;
int i;
for (i = 1; i <= N; i++)
{
if (!visited[i] && edge[i][start] == 1)
DFS(i);
}
} int main()
{
int i, j;
int a, b;
scanf("%d%d%d", &N, &M, &K);
for (i = 0; i<M; i++)
{
scanf("%d%d", &a, &b);
edge[a][b] = 1;
edge[b][a] = 1;
}
int temp;
int num;
for (i = 0; i<K; i++)
{
num = 0;
scanf("%d", &temp);
memset(visited, 0, sizeof(visited));
visited[temp] = 1;
for (j = 1; j <= N; j++)
{
if (visited[j] == 0)
{
DFS(j);
num++;
}
}
if (num == 0) printf("0\n");
else printf("%d\n", num - 1);
} }
另外一种方法是使用并查集，将所有有边的结点并入同一个集合，并且在查找父节点时进行路径压缩，保证在一个集合中的所有父节点均指向祖先节点，这时候只要计数祖先节点的数目，就可以找出独立区域的个数。
这个方法是从sunbaigui的博客上学到的，下面是他的算法，我在理解的基础上加了些注释，方便阅读。
#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std; int n;//number of city
int m;//number of edge
int k;//number of query
typedef struct Edge
{
int v;
Edge(int _v) :v(_v){};
}Edge;
typedef struct Node
{
int parent;
}Node;
vector<Node> city;
void InitSet()
{
city.resize(n); // 使用vector的resize方法可以重设vector大小
for (int i = 0; i < n; ++i)
city[i].parent = i; // 初始化每个人的父节点为自己
}
void CompressSet(int top, int x) // 压缩路径的目的是让x的父节点指向top
{
if (city[x].parent != top) // 如果x的父节点不指向top，应当让它指向top，注意x的父节点和之前的路径也要压缩，因此先递归再赋值。
{
CompressSet(top, city[x].parent); // 继续向上压缩
city[x].parent = top; // 调整父节点为top
}
}
int FindSet(int x) // 找到x的父节点，并且在寻找之前压缩路径
{
if (city[x].parent != x) // 如果x的父节点不是自己，说明这个集合不是单个元素的集合，应该把所有元素指向祖先节点。
{
int top = FindSet(city[x].parent); // 先获取集合的祖先节点
CompressSet(top, x); // 压缩x到祖先路径上的所有结点指向top
}
return city[x].parent; // 如果x的父节点为自己，说明这就是祖先节点。
}
void UnionSet(int x, int y)
{
int a = FindSet(x);
int b = FindSet(y);
city[a].parent = b; // 集合的合并即找到二者的父亲，让其中一个的父亲成为另一个结点，因为这里不涉及到计数，因此不必判断集合大小。
} int main()
{
//input
scanf("%d%d%d", &n, &m, &k);
vector<vector<Edge>> edge;
edge.resize(n);
for (int i = 0; i < m; ++i)
{
int a, b;
scanf("%d%d", &a, &b);
a--; b--;
edge[a].push_back(Edge(b));
edge[b].push_back(Edge(a));
}
//
//query
for (int i = 0; i < k; ++i)
{
int q;
scanf("%d", &q);
q--;
InitSet(); // 每个元素都是祖先
for (int u = 0; u < n; ++u)
{
for (int j = 0; j < edge[u].size(); ++j)
{
int v = edge[u][j].v;
if (u != q&&v != q) UnionSet(u, v); // 把v w结点的集合合并
}
}
// 这时候所有结点都加入了集合，但是注意到还没有压缩路径，因此很多结点的父节点并未成功指向祖先节点。
set<int> parentSet;
for (int j = 0; j < n; ++j){
// 为了保证所有结点指向祖先结点，使用FindSet函数来查找parent，而不能直接获取parent成员变量，一定要注意！
parentSet.insert(FindSet(j)); // 由于没有排除被占领的结点自成一个父节点，故要减去它才是真正的所有未占领结点组成的集合。
}
printf("%d\n", parentSet.size() - 2);
}
return 0;
}