PAT: 1013. Battle Over Cities (25)

1013. Battle Over Cities (25)

时间限制

400 ms

内存限制

32000 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

100

AC代码：

//“畅通工程”---最小生成树的变异版本#include<stdio.h>int p[1000]; //存放根节点int spoint[500000], epoint[500000]; //存放highway连接的两点int findRoot(int x) { //并查集if(p[x]==x)return x;else {int tmp = findRoot(p[x]); //压缩查找路径p[x] = tmp;return tmp;}}int main(){    //freopen("in.txt","r",stdin);int n, m , k, i;scanf("%d %d %d",&n,&m,&k);int a, b;for(i=0; i<m; i++) {scanf("%d %d",&a,&b);spoint[i] = a;epoint[i] = b;}int check;while(k--) {for(i=1; i<=n; i++) { //初始化并查集p[i] = i;}scanf("%d",&check);for(i=0; i<m; i++) {if(spoint[i]!=check && epoint[i]!=check) { //对于含有要检查的点的路径的两端点不加入并查集a = findRoot(spoint[i]);b = findRoot(epoint[i]);if(a!=b) {p[b] = a;}}}int count=0;for(i=1; i<=n; i++) {if(p[i] == i && p[i]!=check)count++; //统计除check节点外的连通分量个数}printf("%d\n",count-1);}    return 0;}