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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3479 Accepted Submission(s): 1350
1 #include <queue> 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 char map[2020][2020]; 7 int indegree[2020]; 8 int n; 9 bool Solve() 10 { 11 queue<int> q; 12 for(int i = 0; i < n; i++) 13 { 14 if(indegree[i] == 1) 15 q.push(i); 16 } 17 int num = 0; 18 while(!q.empty()) 19 { 20 int temp = q.front(); 21 q.pop(); 22 num++; 23 for(int i = 0; i < n; i++) 24 { 25 if(map[temp][i]) 26 { 27 indegree[i]--; 28 if(indegree[i] == 0) 29 q.push(i); 30 } 31 } 32 } 33 if(num == n) 34 return false; 35 else 36 return true; 37 } 38 int main() 39 { 40 int t, temp = 1; 41 scanf("%d", &t); 42 while(t--) 43 { 44 scanf("%d", &n); 45 memset(indegree, 0, sizeof(indegree)); 46 for(int i = 0; i < n; i++) 47 { 48 scanf("%s", map[i]); 49 for(int j = 0; j < n; j++) 50 if(map[i][j] == ‘1‘) 51 indegree[j]++; 52 } 53 if(Solve()) 54 printf("Case #%d: Yes\n", temp++); 55 else 56 printf("Case #%d: No\n", temp++); 57 } 58 return 0; 59 }
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 using namespace std; 5 char map[2020][2020]; int indegree[2020]; 6 int main() 7 { 8 int t, temp = 1; 9 scanf("%d", &t); 10 while(t--) 11 { 12 int n; 13 scanf("%d", &n); 14 bool flag = false; 15 memset(indegree, 0, sizeof(indegree)); 16 for(int i = 0; i < n; i++) // 计算每个点入度； 17 { 18 scanf("%s", map[i]); 19 for(int j = 0; j < n; j++) 20 if(map[i][j] == ‘1‘) 21 indegree[j]++; 22 } 23 for(int i = 0; i < n; i++) 24 { 25 int j; 26 for(j = 0; j < n; j++) 27 if(indegree[j] == 0) 28 break; 29 if(j == n) //没有入度为零的点， 存在环。 30 { 31 flag = true; 32 break; 33 } 34 else 35 { 36 indegree[j]--; 37 for(int k = 0; k < n; k++) //与 入度为0的点相接点的入度-1； 38 { 39 if(map[j][k] == ‘1‘ && indegree[k] != 0) 40 indegree[k]--; 41 } 42 } 43 } 44 if(flag) 45 printf("Case #%d: Yes\n", temp++); 46 else 47 printf("Case #%d: No\n", temp++); 48 } 49 return 0; 50 }
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原文地址：http://www.cnblogs.com/fengshun/p/4718866.html
