拓扑排序

https://www.cnblogs.com/31415926535x/p/10367483.html

貌似从来没有敲过拓扑排序的板子,,,记录一下

拓扑排序就是对DAG有向无环图中的边u->v,要求排序出一个点的序列,满足u在v的前面,,

算法的思路是不停的将入度为零的点u放到前面,并且对u能到达的所有点v的入度递减,,循环处理所有的点即可,,期间将所有入度为零的点放在一个队列中,,

板子题

这道题要求对于多种可能的排序输出字典序最小的那种,,用优先队列代替原来的队列就行了,,

  • 注意杭电上不能用万能头文件,而且优先队列的由小到大的写法 priority_queue<int, vector<int>, greater<int> > q;,头文件要加 #include <queue>#include <functional> (一直不知道,,,233,,,
  • 还有好久不练忘记多组数据要记得清零那些数组,,

代码:

//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string.h>
#include <vector>
#include <queue>
#include <functional>
#define aaa cout<<233<<endl;
#define endl '\n'
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int inf = 0x3f3f3f3f;//1061109567
const ll linf = 0x3f3f3f3f3f3f3f;
const double eps = 1e-6;
const double pi = 3.14159265358979;
const int maxn = 1e5 + 5;
const int maxm = 2e5 + 5;
const int mod = 1e9 + 7;
inline ll read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}

//red_book
//l[maxn]为最后排序的结果
vector<int> g[maxn];
int du[maxn], n, m, l[maxn];
bool toposort()
{
    memset(du, 0, sizeof du);
    for(int i = 1; i <= n; ++i)
        for(int j = 0; j < g[i].size(); ++j)
            ++du[g[i][j]];
    int tot = 0;
    priority_queue<int, vector<int>, greater<int> > q;//按字典序最小的排序时
    //queue<int> q;
    for(int i = 1; i <= n; ++i)
        if(!du[i])
            q.push(i);
    while(!q.empty())
    {
        int x = q.top(); 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;
    else        return 0;
}
int main()
{
//    freopen("233.in" , "r" , stdin);
//    freopen("233.out" , "w" , stdout);
//    ios_base::sync_with_stdio(0);
//    cin.tie(0);cout.tie(0);
    int u, v;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        for(int i = 0; i <= n; ++i)g[i].clear();
        for(int i = 0; i <= n; ++i)du[i] = l[i] = 0;
        for(int i = 1; i <= m; ++i)
        {
            scanf("%d%d", &u, &v);
            g[u].push_back(v);
        }
        toposort();
        for(int i = 0; i < n - 1; ++i)printf("%d ", l[i]);printf("%d\n", l[n - 1]);
    }
    return 0;
}

(end)

posted @ 2019-02-12 22:59  31415926535x  阅读(235)  评论(0编辑  收藏  举报