大意: 给定$m$个有序对$(a,b)$, 求构造一个$n$排列, 满足$m$个对中$a$均排在$b$前, 且$1$尽量靠前, 在$1$尽量靠前的前提下$2$尽量靠前,....以此类推.
答案为反向拓排的最大字典序.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘\n‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head #ifdef ONLINE_JUDGE const int N = 1e6+10; #else const int N = 111; #endif int n, m; int x[N], y[N], ans[N]; int deg[N]; vector<int> g[N]; priority_queue<int> q; void work() { scanf("%d%d", &n, &m); REP(i,1,n) g[i].clear(),deg[i]=0; REP(i,1,m) { scanf("%d%d", x+i, y+i); g[y[i]].pb(x[i]),++deg[x[i]]; } REP(i,1,n) if (!deg[i]) q.push(i); int cnt = 0; while (q.size()) { int u = q.top(); q.pop(); ans[++cnt] = u; for (int v:g[u]) { if (!--deg[v]) q.push(v); } } if (cnt!=n) return puts("Impossible!"),void(); PER(i,1,n) printf("%d ", ans[i]);hr; } int main() { int t; scanf("%d", &t); while (t--) work(); }
原文:https://www.cnblogs.com/uid001/p/10815331.html