题目大意是这样的,银行里面账户之间会互相转账,只有入账而没有出账的账户被称为安全账户,若另一个账户能够通过
转账路径到达安全账户,那么此账户也是安全账户。输入m个账户,和n条转账关系,输出安全账户的序号。
#include<iostream>
#include<list>
#include<iterator>
#include<stack>
using namespace std;
int main()
{
int m, n;
cin >> m >> n;
int input[100][2];
int matrix[100][100];
for (int i = 0; i < 100; i++)
{
for (int j = 0; j < 100; j++)
{matrix[i][j] = 0; }
}
for (int i = 1; i <=n;i++)
{
cin >> input[i][0] >> input[i][1];
matrix[input[i][0]][input[i][1]] = 1;
}
list<int> temp;
for (int i = 1; i <= m; i++)//先找到只入不出的终节点
{
int sum = 0;
for (int j = 1; j <= m; j++)
sum += matrix[i][j];
if (sum == 0)
temp.push_back(i);
}
//stack<int> safety;
for (int i = 1; i <= m; i++)
{
if (find(temp.begin(), temp.end(), i) != temp.end())
{
continue;
}
else
{
int b[100] = { 0 };
b[i] = 1;//作为此条路径上账户的标记
int index = input[i][1];//下一个账户
int q = 0;
while (1)
{
if (b[index] == 1)//如果下一个账户已经出现过,跳出循环
{
q = 1;
break;
}
else
b[index] = 1;
if (find(temp.begin(), temp.end(), index) != temp.end())
{
break;
}
else
{
index = input[index][1];
}
}
if (q == 1)
break;
temp.push_back(i);
}
}
list<int>::iterator iter;
for (iter=temp.begin(); iter!=temp.end(); iter++)
{
cout<<*iter << " ";
}
system("pause");
return 0;
}