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.
代码如下:
#include<bits/stdc++.h>
using namespace std;
#define size 505
typedef int U[size];
typedef struct
{
int x;
int y;
}Road;
void Init(U& S,int n)
{
for (int i = 0; i < n; i++)S[i] = -1;
}
int Find(U S,int x)
{
while (S[x] >= 0)x = S[x];
return x;
}
void MergebWeight(U& S, int x, int y)
{
int i = Find(S, x), j = Find(S, y);
if (i != j)
{
int temp = S[i] + S[j];
if (S[i] < S[j])
{
S[j] = i;
S[i] = temp;
}
else
{
S[i] = j;
S[j] = temp;
}
}
}
int CollapsingFind(U S, int i)
{
int k, temp;
for (k = i; S[k] >= 0; k = S[k]);
while (i != k)
{
temp = S[i]; S[i] = k; i = temp;
}
return k;
}
int main()
{
int N, M;
cin >> N >> M;
U f;
Init(f, N);
vector<Road>r;
for (int i = 0; i < M; i++)
{
int a, b;
cin >> a >> b;
Road t;
t.x = a; t.y = b;
r.push_back(t);
MergebWeight(f, a, b);
}
map<int, int>mp;
int sum = 0;
for (int i = 0; i < N; i++)
{
int temp = CollapsingFind(f, i);
if (!mp.count(temp))
{
mp[temp] = 1;
sum++;
}
}
int K, num;
cin >> K;
int vis[505] = { 0 };
for (int i = 0; i < K; i++)
{
Init(f, N);
cin >> num;
vis[num] = 1;
for (int i = 0; i < M; i++)
{
if (vis[r[i].x] == 1 ||vis[r[i].y] == 1)
continue;
MergebWeight(f, r[i].x, r[i].y);
}
mp.clear();
int cnt = 0;
for (int i = 0; i < N; i++)
{
int temp = CollapsingFind(f, i);
if (!mp.count(temp))
{
mp[temp] = 1;
cnt++;
}
}
if (cnt - 1 == sum || cnt == sum)
printf("City %d is lost.\n", num);
else
printf("Red Alert: City %d is lost!\n", num);
sum = cnt;
}
if (K >= N)
printf("Game Over.\n");
return 0;
}