1013 Battle Over Cities

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#图论 #PTA #算法

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本文介绍了一种算法，用于计算当某个城市被敌方占领后，为了保持剩余城市间的连通性，需要修复多少条高速公路。该算法通过广度优先搜索（BFS）来确定新增边的数量。

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:

Sample Output:

1
0
0

题目大意

给你一个无向图,保证初始任意两点可相互抵达。求失去任意一点后,需要至少添加几条路,让剩下的点依旧可以相互抵达

思路

对摧毁的点做标记,判断需要BFS几次才能让所有点被扫描到，结果为BFS扫描次数减一

C/C++

#include<bits/stdc++.h>
using namespace std;
vector<int> road[1000];
void findRoad(int now,bool apr[]);
int main()
{
    int N,M,K,a,b;
    cin >> N >> M >> K;
    while (M--){
        cin >> a >> b;
        road[a].push_back(b);
        road[b].push_back(a);
    }
    while (K--){
        bool apr[1000]={false};
        b = 0;
        cin >> a;
        apr[a] = true;
        for(int z=1;z<=N;z++) {
            if(!apr[z]){
                b++;
                findRoad(z,apr);
            }
        }
        cout << b-1 << endl;
    }
    return 0;
}
void findRoad(int now,bool apr[])
{
    if(apr[now]) return;
    apr[now] = true;
    for(int x:road[now]) findRoad(x,apr);
}