1013. Battle Over Cities (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 31 21 31 2 3

Sample Output

10
0
这个题目实际上也就是在去除某个节点后，计算剩下的连通集个数n,然后需要修复的公路的数目为n-1。可以通过dfs或者bfs来对剩余的图进行遍历，这里使用dfs来进行遍历。
#include <iostream>#include <queue>using namespace std;int num_city,num_highway,k;int road[1000][1000]={0};int visited[1000]={0};void dfs(int city);int main(){    queue<int> q;    cin >>num_city >>num_highway >>k;    for(int i=0;i<num_highway;i++)    {        int a,b;        cin >>a >>b ;        road[a][b]=road[b][a]=1;    }    for(int i=0;i<k;i++)    {        int a;        cin >>a;        q.push(a);    }    int cnt=-1;    for(int i=0;i<k;i++)    {        int occupied=q.front();        q.pop();        visited[occupied]=1;        for(int j=1;j<num_city+1;j++)        {            if(visited[j]!=1)            {                visited[j]=1;                dfs(j);                cnt++;            }        }        cout <<cnt <<endl;        cnt=-1;        for(int i=1;i<num_city+1;i++)            visited[i]=0;    }    return 0;}void dfs(int city){    for(int i=1;i<num_city+1;i++)    {        if(visited[i]!=1 && road[city][i]==1)        {            visited[i]=1;            dfs(i);        }    }}