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 = 50; vector<int> g[maxn]; int du[maxn], n, m, L[maxn]; int topsort() { memset(du, 0, sizeof(du)); int flag = 0; for (int i=0; i<n; i++) { for (int j=0; j<g[i].size(); j++) { du[g[i][j]]++; } } int tot = 0; int ct = 0; queue<int> Q; for (int i=0; i<n; i++) { if (!du[i]) { Q.push(i); ct++; } } if (ct > 1) flag = -1;//无序 while (!Q.empty()) { ct = 0; int x = Q.front(); Q.pop(); L[tot++] = x; for (int j=0; j<g[x].size(); j++) { int t = g[x][j]; du[t]--; if (!du[t]) { Q.push(t); ct ++; } } if (ct > 1) flag = -1;//无序 } if (flag == -1) {//无序情况下优先判断是否冲突 if (tot != n) return 0;//有环 else return -1; } if (tot == n) return 1; return 0;//有环 } int main() { while (cin>>n>>m && (m||n)) { int flag = 0; memset(g, 0, sizeof(g)); for (int i=0; i<m; i++) { string str; cin>>str; if (flag) continue; int a = str[0] - 'A'; int b = str[2] - 'A'; g[a].push_back(b); int ans = topsort(); if (ans == 1) {//有序 cout<<"Sorted sequence determined after "<<i+1<<" relations: "; for (int i=0; i<n; i++) { cout<<char(L[i]+'A'); } cout<<"."<<endl; flag = 1; } if (ans == 0) {//环 ,冲突 cout<<"Inconsistency found after "<<i+1<<" relations."<<endl; flag = 1; } } if (!flag) {//无序 cout<<"Sorted sequence cannot be determined."<<endl; } } return 0; }