算法模板之拓扑排序(HDU 3342 Legal or Not)

模板总结归纳:

//拓扑排序
//O(|V| + |E|)

const int maxn = 1e5+5;
vector <int> g[maxn];
int du[maxn], n, m, L[maxn];  // L储存拓扑排序结果 

bool topsort(){
	memset(du, 0, sizeof(du));
	for(int i = 0; i < n; i++)
		for(int j = 0; j < g[i].size(); j++)
			du[g[i][j]]++;
	int tot = 0;
	queue <int> Q;
	for(int i = 0; i < n; i++)
		if(!du[i]) Q.push(i);
	while(!Q.empty()){
		int x = Q.front(); Q.pop();
		L[tot++] = x;
		for(int j = 0; j < g[x].size(); j++){
			int t = g[x][j];
			du[t]--;
			if(!du[t])  Q.push(t);
		}
	}
	if(tot == n)   return 1;
	return 0;
}

实战模板题 :HDU 3342 Legal or Not

#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<cstring>

using namespace std;

const int maxn = 110;
vector <int> g[maxn];
int du[maxn], n, m, L[maxn];

bool topsort(){
	memset(du, 0, sizeof(du));
	for(int i = 0; i < n; i++)
		for(int j = 0; j < g[i].size(); j++)
			du[g[i][j]]++;
	int tot = 0;
	queue <int> Q;
	for(int i = 0; i < n; i++)
		if(!du[i]) Q.push(i);
	while(!Q.empty()){
		int x = Q.front(); Q.pop();
		L[tot++] = x;
		for(int j = 0; j < g[x].size(); j++){
			int t = g[x][j];
			du[t]--;
			if(!du[t])  Q.push(t);
		}
	}
	if(tot == n)   return 1;
	return 0;
}

int main()
{
	while(cin >> n >> m)
	{
		if(n == 0 && m == 0) break;
		int a, b;
		for(int i = 0; i < n; i++) g[i].clear();
		while(m--)
		{
			cin >> a >> b;
			g[a].push_back(b);
		}
		if(topsort()) cout << "YES" << endl;
		else cout << "NO" << endl;
	}
	return 0;
}
    原文作者:拓扑排序
    原文地址: https://blog.csdn.net/lonelydoge/article/details/80101851
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