PAT(甲级)1013
2015-09-23 16:35
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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 3 1 2 1 3 1 2 3
Sample Output
1 00
#include <iostream>
#include <vector>
#define SIZE 1003
using namespace std;
struct Node{
Node(int i){
number = i;
Next = NULL;
}
int number;
Node *Next;
};
struct City{
// int number;
City(){
visited = false;
flag = false;
Next = NULL;
}
bool visited;
bool flag;
Node *Next;
Node *last;
void AddNode(int i);
};
void City::AddNode(int i){
if(last != NULL){
last->Next = new Node(i);
last = last->Next;
}else{
Next = new Node(i);
last = Next;
}
}
City C[SIZE];
void DFS(City C[],int &rootnumber){
City *pcity = &C[rootnumber];
pcity->visited = true;
Node *pNode = pcity->Next;
while(pNode != NULL){
int number = pNode->number;
if(!C[number].visited && !C[number].flag)
DFS(C,pNode->number);
pNode = pNode->Next;
}
}
void Reset(City C[], int n){
for(int i = 1;i<=n;i++){
C[i].visited = false;
C[i].flag = false;
}
}
void Display(City C[],int N)
{
for(int i=1;i<N;i++){
cout <<i <<' ';
City *pcity = &C[i];
Node *pNode = pcity->Next;
while(pNode != NULL){
cout <<pNode->number <<' ';
pNode = pNode->Next;
}
cout <<endl;
}
}
int main()
{
int N,M,K;
int c1,c2;
// freopen("test.txt","r",stdin);
scanf("%d%d%d",&N,&M,&K);
for(int i=1;i<=M;i++){
scanf("%d%d",&c1,&c2);
C[c1].AddNode(c2);
C[c2].AddNode(c1);
}
// Display(C,N);
for(int i=0;i<K;i++){
scanf("%d",&c1);
Reset(C,N);
C[c1].flag = true;
int count = 0;
for(int j=1;j<=N;j++){
if(C[j].flag)
continue;
else{
if(!C[j].visited){
count++;
DFS(C,j);
}
}
}
printf("%d\n",count-1);
}
// fclose(stdin);
return 0;
}
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