【解析】1013 Battle Over Cities (25 分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

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.
I
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.

IInput 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.

IOutput Specification:
For each of the K cities, output in a line the number of highways need to be repaired if that city is lost.

ISample Input:
3 2 3
1 2
1 3
1 2 3

Sample Output:
1
0
0

题意：n座城，m条路，k个沦陷的城市，要求每个城市之间都要有通信，求某城市沦陷后至少修几条路才能达到要求。

分析：经典连通块问题。计算城市网中有多少连通块，连通块个数-1就是要修的道路数(连通块彼此相连，则所有城市相连)。

解法：DFS解法与拓扑排序解法皆可。先写了DFS解法，拓扑排序解法明天更新~

注意：如果用cin、cout输出，最好加上ios::sync_with_stdio(false);这句代码，可以减少大概100ms的时间规模(比scanf、printf还快)。

关闭流同步前：

关闭流同步后：

代码一：DFS解法

#include<bits/stdc++.h>
using namespace std;int n, m, k, lost_city, G[1010][1010], vis[1010];void dfs(int u) {for(int v = 1; v <= n; v++) if(G[u][v] == 1 && vis[v] == 0 && v != lost_city) {vis[v] = 1;dfs(v);
//          vis[v] = 0;        求连通块时无回溯 }
}int main() {ios::sync_with_stdio(false);cin >> n >> m >> k;for(int i = 0; i < m; i++) {int a, b; cin >> a >> b;G[a][b] = G[b][a] = 1;}for(int i = 0; i < k; i++) {cin >> lost_city;memset(vis, 0, sizeof(vis));  //数组初始化int num = -1;       //一定有一块连通块，因此初值-1 for(int j = 1; j <= n; j++) {   //逐个点遍历 if(vis[j] == 0 && j != lost_city) {  //没遍历过且不等于x(x城已消失) vis[j] = 1; dfs(j);num++;}cout << num << '\n';} return 0;
}

代码二：拓扑排序解法（待更）

耗时：

【解析】1013 Battle Over Cities (25 分)_31行代码AC相关推荐

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【解析】1013 Battle Over Cities (25 分)_31行代码AC

立志用最少的代码做最高效的表达

代码一：DFS解法

代码二：拓扑排序解法（待更）

耗时：

【解析】1013 Battle Over Cities (25 分)_31行代码AC相关推荐

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