Luogu3243 [HNOI2015]菜肴制作 （拓扑排序）

题面毒人，其实就是叫你反图跑拓扑

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long

//#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;
#include<queue>

const int N = 100007;

struct Edge{
    int nxt, pre;
}e[N << 1];
int head[N], cntEdge;
inline void add(int u, int v){
    e[++cntEdge] = (Edge){ head[u], v}, head[u] = cntEdge;
}

int in[N];
priority_queue<int>q;
int ans[N];
int main(){
    int Tasks;
    io >> Tasks;
    int flag = 0;
    while(Tasks--){
        int n, m;
        io >> n >> m;
        if(flag){
            R(i,0,n){
                head[i] = in[i] = 0;
            }
            cntEdge = 0;
        }
        
        R(i,1,m){
            int u, v;
            io >> u >> v;
            add(v, u);
            ++in[u];
        }
        
        int tot = 0;
        R(i,1,n){
            if(!in[i]){
                q.push(i);
            }
        }
        while(!q.empty()){
            int u = q.top();
            q.pop();
            ans[++tot] = u;
            for(register int i = head[u]; i; i = e[i].nxt){
                int v = e[i].pre;
                --in[v];
                if(!in[v]){
                    q.push(v);
                }
            }
        }
        
        if(tot != n){
            printf("Impossible!\n");
            flag = 1;
            continue;
        }
        
        nR(i,n,1){
            printf("%d ", ans[i]);
        }
        putchar('\n');
        flag = 1;
    }
    
    return 0;
}

Luogu3243 [HNOI2015]菜肴制作 （拓扑排序）

原文：https://www.cnblogs.com/bingoyes/p/11229783.html