| 题目描述 |
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 city~1~-city~2~ and city~1~-city~3~. Then if city~1~ is occupied by the enemy, we must have 1 highway repaired, that is the highway city~2~-city~3~.
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
| 解题思路 |
题目大意为，已知n个城市之间有m条相连的道路，当消除一个城市及其相连的道路，问若要将其他剩余的城市再次变成连通的话，至少需要增加多少条道路
- 首先是一个连通图的问题，即去掉一个结点之后，若要再次连通至少需要增加多少条路线 ．对于互相分立的连通分量数目为a，那么至少需要增加a-1条路线，便可将其连通
- 需要注意的是，可能结点数为１，去掉该结点后，连通分量数为０，若再减１，将是不符合条件的
| 代码设计 |
//部分代码
//zhicheng
//Jul.30,2018
int v[1010][1010];
bool visit[1010];
void dfs(int node,int n)
{
visit[node]=true;
for(int i=1;i<=n;i++) if(visit[i]==false&&v[node][i]==1) dfs(i,n);
}
int main()
{
int n,m,k,a,b;
scanf("%d %d %d",&n,&m,&k);
for(int i=0;i<m;i++) {scanf("%d %d",&a,&b);v[a][b]=v[b][a]=1;}
for(int i=0;i<k;i++)
{
int cnt=0;
fill(visit,visit+1010,false);
scanf("%d",&a);
visit[a]=true;
for(int j=1;j<=n;j++) if(visit[j]==false){dfs(j,n);cnt++;}
printf("%d\n",cnt?cnt-1:cnt);
}
return 0;
}
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