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 Specification:
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 Specification:
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
這道題目是對并查集的一種典型應用，可以轉化為：每次占領一個城市，并查集連接配接時便略過該城市，最後檢視有幾個互不相幹的集合n，
便說明了至少還需要n-1條路，代碼如下：
import java.util.*;public classBattleOverCities2 {static int[] weight ;static int[] parent;static intnums ;public static voidmain(String args[])
{
Scanner scanner= newScanner(System.in);int cities =scanner.nextInt();int roads =scanner.nextInt();int checked =scanner.nextInt();int[][] road = new int[roads][2];for(int i = 0 ; i < roads ; i ++)
{
road[i][0] = scanner.nextInt(); //road用來記錄有多少條道路
road[i][1] =scanner.nextInt();
}int[] check = new int[checked] ;for(int i = 0 ; i < checked ; i ++)
{
check[i]= scanner.nextInt(); //check[]用來記錄被占領的城市
}
scanner.close();for(int i = 0 ; i < checked ; i ++)
{
init(cities);
solve(road,check[i]);
System.out.println(nums-2);
}
}public static void init(intcities)
{
parent= new int[cities+1] ;
nums=cities ;
weight= new int[cities+1] ;for(int i = 1 ; i < cities + 1 ; i ++)
{
parent[i]=i ;
weight[i]= 1;
}
}public static int find(intcitiy)
{while(citiy !=parent[citiy])
{
citiy=parent[parent[citiy]];
}returncitiy ;
}public static void union(int[] road )
{int citiy1 = find(road[0]) ;int citiy2 = find(road[1]) ;if(citiy1 == citiy2) return;if(weight[citiy1] >weight[citiy2])
{
parent[citiy2]=citiy1 ;
weight[citiy1]+=weight[citiy2] ;
}else{
parent[citiy1]=citiy2 ;
weight[citiy2]+=weight[citiy1] ;
}
nums--;
}public static void solve(int[][] road , intcheck)
{for(int i = 0 ; i < road.length ; i ++)
{if(road[i][0] == check || road[i][1] == check) //當需要連接配接的城市中有一座被占領，則掠過這條路
continue;
union(road[i]) ;
}
}
}
還有一道題目：
1021 Deepest Root (25 分)
A graph which is connected and acyclic can be considered a tree. The hight of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤104) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N−1 lines follow, each describes an edge by given the two adjacent nodes' numbers.
Output Specification:
For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print Error: K components where K is the number of connected components in the graph.
Sample Input 1:
5
1 2
1 3
1 4
2 5
Sample Output 1:
3
4
5
Sample Input 2:
5
1 3
1 4
2 5
3 4
Sample Output 2:
Error: 2 components
這道題目需要求圖的生成樹的最大深度是多少，首先需要判斷圖是否都相連，這裡用到了并查集，如果互不相連的集合數目大于1，則傳回互不相連的
個數，如果為1，則使用dfs找到生成樹的最大深度，代碼如下：
import java.util.*;public classDeepestRoot {static intnums ;public static voidmain(String args[])
{
Scanner scanner= newScanner(System.in);
nums=scanner.nextInt() ;int[] nodes = new int[nums + 1] ;int[] weight = new int[nums + 1] ;
Map> map = new HashMap<>() ;for(int i = 1 ; i < nums + 1 ; i ++)
{
nodes[i]=i ;
weight[i]= 1;
}int t =nums ;for(int i = 1 ; i < t ; i ++)
{int p =scanner.nextInt() ;int q =scanner.nextInt() ;
union(p,q,nodes,weight);if(!map.containsKey(p))
map.put(p,new ArrayList()) ;if(!map.containsKey(q))
map.put(q,new ArrayList()) ;
map.get(p).add(q) ;
map.get(q).add(p) ;
}
scanner.close();if(nums != 1)
{
System.out.println("Error: "+nums+" components");
}else{int max = 0;
Map> mp = new HashMap<>() ;for(int i = 1 ; i < t + 1 ; i ++)
{if(map.get(i).size() == 1)
{int c = dfs(map,i,0,new int[t+1],0) ;if(c >max)
{
mp.put(c,new ArrayList<>()) ;
mp.get(c).add(i);
max=c ;
}else if(c ==max)
mp.get(max).add(i) ;
}
}
Collections.sort(mp.get(max));
System.out.print(mp.get(max).get(0));for(int i = 1 ; i < mp.get(max).size() ; i ++)
{
System.out.println();
System.out.print(mp.get(max).get(i));
}
}
}public static int dfs(Map> map,int q,int count,int[] checked,intmax)
{
checked[q]= 1;int p = 0;for(int i = 0 ; i < map.get(q).size() ; i ++)
{int t =map.get(q).get(i);if(checked[t] == 1) continue;int l = dfs(map,t,count+1,checked,max) ;if(max < l) max =l ;
p++;
}if(p == 0) returncount ;returnmax ;
}public static int find(int n,int[] parent)
{while(parent[n] !=n)
{
n=parent[n] ;
}returnn ;
}public static void union(int p,int q,int[] nodes,int[] weight)
{int n =find(p,nodes) ;int m =find(q,nodes) ;if(n == m ) return;if(weight[n] >weight[m])
{
nodes[m]=n ;
weight[n]+=weight[m] ;
nums--;
}else{
nodes[n]=m ;
weight[m]+=weight[n] ;
nums--;
}
}
}