L2-013. 红色警报
时间限制
400 ms
内存限制
65536 kB
代码长度限制
8000 B
判题程序
Standard
作者
陈越
战争中保持各个城市间的连通性非常重要。本题要求你编写一个报警程序,当失去一个城市导致国家被分裂为多个无法连通的区域时,就发出红色警报。注意:若该国本来就不完全连通,是分裂的k个区域,而失去一个城市并不改变其他城市之间的连通性,则不要发出警报。
输入格式:
输入在第一行给出两个整数N(0 < N <=500)和M(<=5000),分别为城市个数(于是默认城市从0到N-1编号)和连接两城市的通路条数。随后M行,每行给出一条通路所连接的两个城市的编号,其间以1个空格分隔。在城市信息之后给出被攻占的信息,即一个正整数K和随后的K个被攻占的城市的编号。
注意:输入保证给出的被攻占的城市编号都是合法的且无重复,但并不保证给出的通路没有重复。
输出格式:
对每个被攻占的城市,如果它会改变整个国家的连通性,则输出“Red Alert: City k is lost!”,其中k是该城市的编号;否则只输出“City k is lost.”即可。如果该国失去了最后一个城市,则增加一行输出“Game Over.”。
输入样例:5 4 0 1 1 3 3 0 0 4 5 1 2 0 4 3输出样例:
City 1 is lost. City 2 is lost. Red Alert: City 0 is lost! City 4 is lost. City 3 is lost. Game Over.
# include <iostream> # include <numeric> # include <algorithm> # include <functional> # include <list> # include <map> # include <set> # include <stack> # include <deque> # include <queue> # include <vector> # include <ctime> # include <cstdlib> # include <cmath> # include <string> # include <cstring> using namespace std; int pre[505]; struct node { int x, y; }; void fun() { for(int i = 0; i < 505; i++) { pre[i] = i; } } int Find(int x) { if(x != pre[x]) { pre[x] = Find(pre[x]); } return pre[x]; } void join(int x, int y) { int fx = Find(x); int fy = Find(y); if(fx != fy) { pre[fy] = fx; } } int main(int argc, char *argv[]) { int n, m; cin >> n >> m; fun(); struct node s[5005]; for(int i = 0; i < m; i++) { cin >> s[i].x >> s[i].y; join(s[i].x, s[i].y); } int k; cin >> k; int lost[505] = {0}; int c1 = 0; for(int i = 0; i < n; i++) { if(pre[i] == i) { c1++; } } int i; for(i = 0; i < k; i++) { int city; cin >> city; //每次的初始化 fun(); int c2 = 0; lost[city] = 1; for(int j = 0; j < m; j++) { //因为是合并,所以要求指向和被指向都要为0才行 if(!lost[s[j].x] && !lost[s[j].y]) { join(s[j].x, s[j].y); } } //计算更新数据后的城市根结点的个数 for(int p = 0; p < n; p++) { if(!lost[p] && pre[p] == p) { c2++; } } if(c1 == c2 || c1 == c2 + 1) { printf("City %d is lost.\n",city); } else { printf("Red Alert: City %d is lost!\n",city); } c1 = c2; } if(k == n) { printf("Game Over.\n"); } return 0; }