pku 1094 Sorting It All Out 拓扑排序

#include <iostream>
#include <stack>
using namespace std;
int edge[27][27]; //图的邻接矩阵
int in_degree[27]; //顶点的入度
bool flagDegree[27];
char strRet[27];
int n, m;

//拓扑排序
//-1: 继续
//0: 完成排序
//1: 有回路
int TopoOrder()
{
stack<int> s;
memset(in_degree, 0, sizeof(in_degree));
memset(strRet, 0, sizeof(strRet));
memset(flagDegree, 0, sizeof(flagDegree));

for (int i = 1; i <= 26; ++i)
for (int j = 1; j <= 26; ++j)
{
if(edge[i][j])
{
flagDegree[i] = flagDegree[j] = 1;
in_degree[j]++;
}
}

for (int i = 1; i <= 26; ++i)
if(flagDegree[i] && !in_degree[i])
s.push(i);

int cnt = 0;
bool flagCannotDetermined = 0;
while(!s.empty())
{
if(s.size() > 1)
flagCannotDetermined = 1;

int temp = s.top();
strRet[cnt++] = 'A'+temp-1;
s.pop();

for(int i = 1; i <= 26; ++i)
if(edge[temp][i] && (--in_degree[i]) == 0)
s.push(i);
}
strRet[cnt] = '/0';

for(int i = 1; i <= 26; ++i)
if(flagDegree[i] && in_degree[i])
return 1;

if(flagCannotDetermined)
return -1;

if(cnt == n)
return 0;

return -1;
}

int main()
{
char str[4];
int i;
while (scanf("%d%d", &n, &m) && !(n==0&&m==0))
{
memset(edge, 0, sizeof(edge));

bool flagOver = 0, flagCircle = 0, flagNotDetermined = 0;
int index;

for (i = 1; i <= m; ++i)
{
cin >> str;
edge[str[0]-'A'+1][str[2]-'A'+1] = 1;

if(flagCircle || flagNotDetermined || flagOver)
continue;

int ret = TopoOrder();
if(ret == 0)
{
flagOver = 1;
index = i;
}
else if(ret == 1)
{
flagCircle = 1;
index = i;
}
if(i == m && ret == -1)
flagNotDetermined = 1;
}
if(flagCircle)
printf("Inconsistency found after %d relations./n", index);
else if(flagNotDetermined)
printf("Sorted sequence cannot be determined./n");
else
printf("Sorted sequence determined after %d relations: %s./n", index, strRet);
}
return 0;
}

//这组数据拯救了我,贡献出来
/*
6 8
A<B
A<C
B<D
C<E
D<E
E<D
D<F
E<F
Right Answer: Inconsistency found after 6 relations.
Wrong Answer: Sorted sequence cannot be determined.
原因是: 判断有回路的优先级比判断不可排序的优先级高.
判断不可排序就会返回去录入新的一组值.

我是这样才行的.
if(s.size() > 1)
flagCannotDetermined = 1;
而不可以直接return -1;//-1代表继续.

*/