1013 Battle Over Cities (DFS+连同分量)

蔚落 2022-03-22 13:40 140阅读 0赞

1013 Battle Over Cities （25 point(s)）

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 Mlines 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
    0
    0

解题思路:

遍历图,得到连通分量的个数,答案就是连通分量的个数-1(边数=结点数-1）

这里利用广度优先会超时,故使用深度优先遍历,同时注意要用scanf和printf输入输出数据,否则也会超时

#define _CRT_SECURE_NO_WARNINGS
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <string.h>
    using namespace std;
    
    class Graph {
    public:
    	int n;
    	vector<int>* edges;
    	bool* visied;
    public:
    	Graph(int _n) {
    		n = _n;
    		edges = new vector<int>[n];
    		visied = new bool[n];
    	}
    	void gInsert(int x, int y) {
    		edges[x].push_back(y);
    		edges[y].push_back(x);
    	}
    	void DFS(int tnode,int occity) {
    		visied[tnode] = true;
    		for (int adj_node : edges[tnode]) {
    			if (!visied[adj_node] && adj_node != occity) {
    				DFS(adj_node,occity);
    			}
    		}
    	}
    };
    
    int main() {
    	int N, M, K;
    	scanf("%d %d %d", &N, &M, &K);
    	Graph battleMap(N + 1);
    	for (int i = 0; i < M; ++i) {
    		int nx, ny;
    		scanf("%d %d", &nx, &ny);
    		battleMap.gInsert(nx, ny);
    	}
    
    	for (int i = 0; i < K; ++i) {
    		memset(battleMap.visied, 0, N + 1);
    		int occity;
    		scanf("%d", &occity);
    		int index = 0;
    		for (int j = 1; j <= N; ++j) {
    			if(j == occity || battleMap.visied[j]) continue;
    			battleMap.DFS(j, occity);
    			index++;
    		}
    		cout << index-1 << endl;
    	}
    
    	system("PAUSE");
    	return 0;
    }