题目链接

BZOJ2535

题解

航班之间的关系形成了一个拓扑图
而且航班若要合法，应尽量早出发
所以我们逆拓扑序选点，能在后面出发的尽量后面出发，不会使其它点变得更劣，容易知是正确的

第二问只需枚举航班\(x\)，拓扑排序时忽视\(x\)，最后无法选点时就是\(x\)最早的时间

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (register int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define res register
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 2005,maxm = 20005,INF = 1000000000;
inline int read(){res int out = 0,flag = 1; res char c = getchar();while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}return out * flag;
}
int n,m,inde[maxn],K[maxn],val[maxn],ans[maxn];
int h[maxn],ne;
struct EDGE{int to,nxt;}ed[maxm];
inline void build(int u,int v){ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;inde[v]++;
}
priority_queue<cp> q;
inline void solve1(){REP(i,n){val[i] = inde[i];if (!val[i]) q.push(mp(K[i],i));}cp u;for (int i = n; i; i--){u = q.top(); q.pop();ans[i] = u.second;Redge(u.second){if (!(--val[to = ed[k].to])) q.push(mp(K[to],to));}}REP(i,n){printf("%d",ans[i]);if (i < n) putchar(' ');}puts("");
}
inline int solve2(int x){while (!q.empty()) q.pop();REP(j,n) val[j] = inde[j]; val[x] = n;REP(j,n) if (!val[j]) q.push(mp(K[j],j));cp u;for (res int j = n; j; j--){if (q.empty()) return j;u = q.top(); q.pop();if (u.first < j) return j;Redge(u.second) if (!(--val[to = ed[k].to])) q.push(mp(K[to],to));}return 1;
}
int main(){n = read(); m = read();REP(i,n) K[i] = read();int a,b;REP(i,m) {a = read(); b = read();build(b,a);}solve1();REP(i,n) printf("%d ",solve2(i));return 0;
}