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  • 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("
    
    ----------
    
    ")
    #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!
    ");
    			flag = 1;
    			continue;
    		}
    		
    		nR(i,n,1){
    			printf("%d ", ans[i]);
    		}
    		putchar('
    ');
    		flag = 1;
    	}
    	
    	return 0;
    }
    

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