一.问题回顾：
Sorting It All Out （POJ - 1094 ，拓扑排序 + 链式向前星存图）
由于之前已经写过拓扑排序解法，这里不在详谈.
二.分析：
Floyd 传闭包
注意：要先判矛盾.
详见代码.
三.代码实现：
#include <bits/stdc++.h>
using namespace std;
const int M = (int)26;
int n, m;
char u, v;
bool flag;
bool dis[M][M];
int Floyed()
{
for(int k = 0; k < n; ++k)
{
for(int i = 0; i < n; ++i)
{
if(dis[i][k])
{
for(int j = 0; j < n; ++j)
{
dis[i][j] |= (dis[i][k] & dis[k][j]);
if(i != j && (dis[i][j] & dis[j][i])) return -1;
}
}
}
}
for(int i = 0; i < n; ++i)
{
for(int j = 0; j < n; ++j)
{
if(i != j && !(dis[i][j] | dis[j][i]))
return 0;
}
}
return 1;
}
int cnt[M];
vector < pair<int, int> > ans;
void print_order()
{
ans.clear();
memset(cnt, 0, sizeof(cnt));
for(int i = 0; i < n; ++i)
{
for(int j = 0; j < n; ++j)
{
if(dis[i][j])
++cnt[j];
}
}
for(int i = 0; i < n; ++i)
ans.push_back(make_pair(cnt[i], i));
sort(ans.begin(), ans.end());
for(auto x : ans)
putchar(x.second + 'A');
}
int main()
{
while(~scanf("%d %d", &n, &m) && (n + m))
{
flag = 0;
memset(dis, 0, sizeof(dis));
for(int i = 1; i <= m; ++i)
{
getchar();
scanf("%c<%c", &u, &v);
if(flag)
continue;
u -= 'A', v -= 'A';
dis[u][v] = 1;
int ans = Floyed();
if(ans == 1)
{
flag = 1;
printf("Sorted sequence determined after %d relations: ", i);
print_order();
printf(".\n");
}
else if(ans == -1)
{
flag = 1;
printf("Inconsistency found after %d relations.\n", i);
}
}
if(!flag)
printf("Sorted sequence cannot be determined.\n");
}
return 0;
}
