題目連接c++
傳送門:https://ac.nowcoder.com/acm/contest/76/Fspa
思路:code
題目的意思就是判斷圖的連通性能夠用可達性矩陣來求,至於圖的存儲能夠用鄰接矩陣來儲存,求出來可達性矩陣後判斷下 a[1] [n] 是否是爲零就行了,不爲零的話說明從1(本身)到 n(teacher)是連通的,爲零的話就說明不是連通的blog
代碼:ci
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 const int MAXN = 60; 5 int n, m, a[MAXN][MAXN]; 6 7 int main() 8 { 9 while (cin >> n >> m) 10 { 11 int x, y; 12 while (m -- ) 13 { 14 cin >> x >> y; 15 a[x][y] = 1; 16 } 17 18 for (int i = 1; i <= n; i ++ ) 19 for (int j = 1; j <= n; j ++ ) 20 { 21 if (a[j][i] == 1) 22 { 23 for (int k = 1; k <= n; k ++ ) 24 if ((a[j][k] + a[i][k]) >= 1) 25 a[i][k] = 1; 26 } 27 } 28 29 if (a[1][n] != 0) puts("Yes"); 30 else puts("No"); 31 memset(a,0,sizeof(a)); 32 } 33 return 0; 34 }