POJ 3687 Labeling Balls (reverse topology)

It is not guaranteed to be the optimal solution for the minimum of the positive each fetch, and the inverse construction edge can get the maximal positive solution at each time.

Code:

#include <iostream> #include <cstdio> #include <cmath> #include <map> #include <queue># include<vector> #include <cstring> #include <algorithm> #define REP (i,a,b) for (int i= (a);i< (b); i+ +) #define REV (I,a,b) for (int i= (a); i>= (b); i--) #define CLR (a,x) memset (a,x,sizeof a) #define INF 0x3f3f3f3ftypedef Long long ll;using namespace std;const int maxn=205;const int Maxm=maxn*maxn;int FIRST[MAXN],NEX[MAXM],V[MAXM],IN[MAXN]    , P[maxn],h[maxn];int n,m,ecnt;bool topsort (int n) {priority_queue<int>q;    memset (h,0,sizeof h);    for (int i=1;i<=n;i++) if (!in[i]) Q.push (i);    int cur=0;        while (!q.empty ()) {int x=q.top (); Q.pop ();        p[x]=n-(cur++);            for (int e=first[x];~e;e=nex[e]) {H[v[e]]=max (h[v[e]],h[x]+1);        if (--in[v[e]]==0) Q.push (V[e]); }} return cur==n;}    void add_ (int a,int b) {v[ecnt]=b;    Nex[ecnt]=first[a]; first[a]=ecnt++;}    int main () {int t,a,b; scanf ("%d",&T);        while (t--) {scanf ("%d%d", &n,&m);        The CLR (first,-1); ecnt=0;        CLR (in,0);            for (int i=0;i<m;i++) {scanf ("%d%d", &a,&b);        Add_ (B,a), in[a]++;        } int ans=topsort (n);            if (!ans) {puts ("-1");        Continue } for (int i=1;i<=n;i++) printf ("%d%c", p[i],i==n? '    \ n ': '); } return 0;}

POJ 3687 Labeling Balls (reverse topology)