UVA - 10305 Ordering Tasks（拓扑排序）

Posted 2020-08-27 SomnusMistletoe

题意：给定优先关系进行拓扑排序。

分析：将入度为0的点加入优先队列，并将与之相连的点入度减1，若又有度数为0的点，继续加入优先队列，依次类推。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int in[MAXN];
vector<int> a[MAXN];
vector<int> ans;
priority_queue<int, vector<int>, greater<int> > q;
int main(){
    int n, m;
    while(scanf("%d%d", &n, &m) == 2){
        if(!n && !m) return 0;
        memset(in, 0, sizeof in);
        ans.clear();
        for(int i = 0; i < MAXN; ++i) a[i].clear();
        while(m--){
            int x, y;
            scanf("%d%d", &x, &y);
            a[x].push_back(y);
            ++in[y];
        }
        for(int i = 1; i <= n; ++i){
            if(in[i] == 0){
                q.push(i);
            }
        }
        while(!q.empty()){
            int t = q.top();
            q.pop();
            ans.push_back(t);
            int len = a[t].size();
            for(int i = 0; i < len; ++i){
                if(--in[a[t][i]] == 0){
                    q.push(a[t][i]);
                }
            }
        }
        int len = ans.size();
        for(int i = 0; i < len; ++i){
            if(i) printf(" ");
            printf("%d", ans[i]);
        }
        printf("\n");
    }
    return 0;
}