題目鏈接

BZOJ2535

題解

航班之間的關(guān)系形成了一個(gè)拓?fù)鋱D
而且航班若要合法，應(yīng)盡量早出發(fā)
所以我們逆拓?fù)湫蜻x點(diǎn)，能在后面出發(fā)的盡量后面出發(fā)，不會(huì)使其它點(diǎn)變得更劣，容易知是正確的

第二問(wèn)只需枚舉航班\(x\)，拓?fù)渑判驎r(shí)忽視\(x\)，最后無(wú)法選點(diǎn)時(shí)就是\(x\)最早的時(shí)間

#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; }

轉(zhuǎn)載于:https://www.cnblogs.com/Mychael/p/9018305.html

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