L2-013. Red Alert
It is important to maintain connectivity between cities during a war. This question asks you to write an alert program that will sound a red alert when the loss of a city causes the country to be divided into multiple unconnected regions. Note: If the country is inherently not fully connected, it is split k regions, and losing a city does not change the connectivity between other cities, don't raise an alarm.
Input format:
The input gives two integers N (0 < N <=500) and M (<=5000) in the first line, which are the number of cities (so the default cities are numbered from 0 to N-1) and the path connecting the two cities number of bars. After M lines, each line gives the number of the two cities connected by a path, separated by 1 space. The captured information is given after the city information, that is, a positive integer K followed by the numbers of the K captured cities.
Note: The input guarantees that the captured city numbers given are valid and without duplication, but it does not guarantee that the given access route is not duplicated.
Output format:
For each captured city, if it would change the connectivity of the entire country, output "Red Alert: City k is lost!", where k is the number of the city; otherwise just output "City k is lost." Can. If the country loses its last city, add a line to output "Game Over.".
Input sample:5 4 0 1 1 3 3 0 0 4 5 1 2 0 4 3Sample output:
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; // each initialization fun(); int c2 = 0; lost[city] = 1; for(int j = 0; j < m; j++) { //Because it is a merge, it is required to point to and be pointed to to be 0. if(!lost[s[j].x] && !lost[s[j].y]) { join(s[j].x, s[j].y); } } //Calculate the number of city root nodes after updating the data 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; }