这道题跟大佬的思路差不多,用一边dfs,或者bfs找到几个连通图,有一点不同就是被占领城市的处理方式不同,把占领的城市设为被访问过显然更快
还有就是scanf,printf比cin,cout快
#include <stdio.h>
#include <algorithm>
using namespace std;
int graph[1010][1010];
int m, n, c;
int visited[1010];
void dfs(int v) {
visited[v] = 1;
for(int i = 1; i <= m; i++) {
if(graph[v][i] == 1 && visited[i] == 0) {
dfs(i);
}
}
}
int main() {
scanf("%d%d%d", &m, &n, &c);
for(int i = 1; i <= n; i++) {
int c1, c2;
scanf("%d%d", &c1, &c2);
graph[c1][c2] = graph[c2][c1] = 1;
}
for(int i = 0; i < c; i++) {
int checkv;
scanf("%d",&checkv);
fill(visited, visited + 1010, 0);
int cnt = 0;
visited[checkv] = 1;
for(int i = 1; i <= m; i++) {
if(visited[i] == 0) {
dfs(i);
cnt++;
}
}
printf("%d\n",cnt-1);
}
return 0;
}
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