求连通分量的问题

使用DFS进行遍历

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 *c it**y*1-

*c*2 and

**i**t**y*c*1-

**i**t**y*c*3. Then if

**i**t**y*c*1 is occupied by the enemy, we must have 1 highway repaired, that is the highway

**i**t**y*c*2-

**i**t**y*c*3.

**i**t**y## Input Specification:

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 Specification:

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
```

## 实现思想

去除一个城市之后求需要修的最少数量的路

就是求连通分量的问题

n个连通分量组成连通图需要n-1条线

## 代码实现

```
#include <cstdio>
#include <algorithm>
using namespace std;
int graph[1010][1010];
bool visit[1010];
int n;
void dfs(int index) {
visit[index] = true;
for(int h = 1; h <= n; h++) {
if(visit[h] == false && graph[index][h] == 1) {
dfs(h);
}
}
}
int main() {
int a, b;
int city, cnt, m, k;
scanf("%d%d%d", &n, &m, &k);
for(int i = 0; i < m; i++) {
scanf("%d%d", &a, &b);
graph[a][b] = graph[b][a] = 1;
}
for(int j = 0; j < k; j++) {
fill(visit, visit+1010, false);
scanf("%d", &city);
cnt = 0;
visit[city] = true;
for(int t = 1; t <= n; t++) {
if(visit[t] == false) {
dfs(t);
cnt++;
}
}
printf("%d\n", cnt-1);
}
return 0;
}
```

**本文链接：**https://blog.zjgsujz.cn/2020/07/21/pat-advanced-1013/**版权声明：**本博客所有文章除特别声明外，均默认采用 许可协议。