1013. Battle Over Cities (25)

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.

3 2 31 21 31 2 3

100

这道题考察最小生成树和集合

用一个二维数组来存储图。

简单考虑的话，需要修的路等于所有的城市数-已有的路径数-2

注意数已有路径数时要避免使最小生成树出现环，即若两个城市已在一个集合里，即使这两个城市中间有路也不需要把这条路数进去。

最后注意set数组在每个case下都要更新一次。

#include 
    
     #define MAX 1000int Graph[MAX][MAX];int set[MAX];int Find(int n);void Union(int a, int b);int main(){	int n, m, k;	scanf("%d %d %d", &n, &m, &k);	for (int i = 1; i <= n; i++) {		for (int j = 1; j <= n; j++)			Graph[i][j] = 0;	}	for (int i = 0; i < m; i++) {		int d1, d2;		scanf("%d %d", &d1, &d2);		Graph[d1][d2] = Graph[d2][d1] = 1;	}	for (int i = 0; i < k; i++) {		int c;		int pathnum = 0;		scanf("%d", &c);		for (int j = 1; j <= n; j++)			set[j] = -1;		for (int j = 1; j <= n; j++) {			if (j == c)				continue;			for (int k = 1; k <= n; k++) {				if (k == c)					continue;				else {					if (Graph[k][j] != 0) {						if (Find(j) != Find(k)) {							Union(k, j);							pathnum++;						}					}				}			}					}		printf("%d\n", n - pathnum - 2);	}	return 0;}int Find(int n){	while (set[n] > 0)		n = set[n];	return n;}void Union(int a, int b){	int roota, rootb;	roota = Find(a);	rootb = Find(b);	if (roota < rootb) {		set[roota] += set[rootb];		set[rootb] = roota;	}	else {		set[rootb] += set[roota];		set[roota] = rootb;	}}