NYOJ的数据水一点,POJ过了是真的过了

/*
拓扑排序模板题:
每次输入都要判断有环与有序的情况,如果存在环路或者已经有序可以输出则跳过下面的输入
判断有序,通过是否在一个以上的入度为0的点,存在则不能有序排列
判断有环,如果拓扑排序完成存在一个有序的排列, 证明无环路
主要判断
1.有序
2.有环
在无序的境况下,优先判断是否有环
有序的情况下,优先判断是否能输出
*/ #include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
using namespace std;
const int maxn = ;
vector<int> g[maxn];
int du[maxn], n, m, L[maxn];
int topsort() {
memset(du, , sizeof(du));
int flag = ;
for (int i=; i<n; i++) {
for (int j=; j<g[i].size(); j++) {
du[g[i][j]]++;
}
}
int tot = ;
int ct = ;
queue<int> Q;
for (int i=; i<n; i++) {
if (!du[i]) {
Q.push(i);
ct++;
}
}
if (ct > ) flag = -;//无序
while (!Q.empty()) {
ct = ;
int x = Q.front();
Q.pop();
L[tot++] = x;
for (int j=; j<g[x].size(); j++) {
int t = g[x][j];
du[t]--;
if (!du[t]) {
Q.push(t);
ct ++;
}
}
if (ct > ) flag = -;//无序
}
if (flag == -) {//无序情况下优先判断是否冲突
if (tot != n) return ;//有环
else return -;
}
if (tot == n) return ;
return ;//有环
}
int main() { while (cin>>n>>m && (m||n)) {
int flag = ;
memset(g, , sizeof(g));
for (int i=; i<m; i++) {
string str;
cin>>str;
if (flag) continue;
int a = str[] - 'A';
int b = str[] - 'A';
g[a].push_back(b);
int ans = topsort();
if (ans == ) {//有序
cout<<"Sorted sequence determined after "<<i+<<" relations: ";
for (int i=; i<n; i++) {
cout<<char(L[i]+'A');
}
cout<<"."<<endl;
flag = ;
}
if (ans == ) {//环 ,冲突
cout<<"Inconsistency found after "<<i+<<" relations."<<endl;
flag = ;
}
}
if (!flag) {//无序
cout<<"Sorted sequence cannot be determined."<<endl;
}
}
return ;
}
05-19 22:33