It is vitally important to has all the cities connected by highways in a war. If a city was occupied by the enemy, all of the highways From/toward that city was closed. We must know immediately if we need to repair all other highways to keep the rest of the cities connected. Given the map of cities which has all the remaining highways marked, you is 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-city 3. Then if City1 is occupied by the enemy, we must has 1 highway repaired, which is the HI Ghway 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 is 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 is the numbers of the cities of Highway. The cities is 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.
#include <stdio.h> #include <stdlib.h> #include <algorithm>using namespace std; #define N 1000int g[n][ N];bool visit[n]={false};int n,m,k;int index;void DFS (int a); int main () {int i,j; scanf ("%d%d%d", &n,&m,&k); Fill (g[0],g[0]+n*n,0); int start,end; for (i=0;i<m;i++) {scanf ("%d%d", &start,&end); G[start][end]=1; G[end][start]=1; } int count=0; for (i=0;i<k;i++) {scanf ("%d", &index); count=0; Fill (visit,visit+n,false); for (j=1;j<=n;j++) {if (j!=index&&visit[j]==false) {DFS (j ); count++; }} printf ("%d\n", count-1); } system ("Pause"); return 0; }void DFS (int a) {int i; if (A==index) return; Visit[a]=true; for (i=1;i<=n;i++) if (visit[i]==false && g[a][i]==1) DFS (i); }
1013. Battle over Cities