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;
}