题意:给定优先关系进行拓扑排序。

分析:将入度为0的点加入优先队列,并将与之相连的点入度减1,若又有度数为0的点,继续加入优先队列,依次类推。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {, , -, , -, -, , };
const int dc[] = {-, , , , -, , -, };
const int MOD = 1e9 + ;
const double pi = acos(-1.0);
const double eps = 1e-;
const int MAXN = + ;
const int MAXT = + ;
using namespace std;
int in[MAXN];
vector<int> a[MAXN];
vector<int> ans;
priority_queue<int, vector<int>, greater<int> > q;
int main(){
int n, m;
while(scanf("%d%d", &n, &m) == ){
if(!n && !m) return ;
memset(in, , sizeof in);
ans.clear();
for(int i = ; i < MAXN; ++i) a[i].clear();
while(m--){
int x, y;
scanf("%d%d", &x, &y);
a[x].push_back(y);
++in[y];
}
for(int i = ; i <= n; ++i){
if(in[i] == ){
q.push(i);
}
}
while(!q.empty()){
int t = q.top();
q.pop();
ans.push_back(t);
int len = a[t].size();
for(int i = ; i < len; ++i){
if(--in[a[t][i]] == ){
q.push(a[t][i]);
}
}
}
int len = ans.size();
for(int i = ; i < len; ++i){
if(i) printf(" ");
printf("%d", ans[i]);
}
printf("\n");
}
return ;
}

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