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  • HDU 5695 Gym Class (拓扑排序)

    题意:略。

    析:当时比赛真是脑残了。。。。一直想什么拓扑模板,其实并不需要么,当时脑子短路了。。。。完全可以直接根据题意写的,用优先队列

    维护一下就好。最近没刷题真是落后了。。。。不说了,太水了。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #define debug() puts("++++");
    #define gcd(a, b) __gcd(a, b)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const LL LNF = 1e16;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 100000 + 10;
    const int mod = 1000000007;
    const int dr[] = {-1, 0, 1, 0};
    const int dc[] = {0, 1, 0, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c){
      return r >= 0 && r < n && c >= 0 && c < m;
    }
    vector<int> G[maxn];
    int in[maxn];
    
    int main(){
      int T;  cin >> T;
      while(T--){
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; ++i)  G[i].clear();
        memset(in, 0, sizeof in);
        for(int i = 0; i < m; ++i){
          int a, b;
          scanf("%d %d", &a, &b);
          G[a].push_back(b);
          ++in[b];
        }
        LL ans = 0;
        int mmin = INF;
        priority_queue<int> pq;
        for(int i = 1; i <= n; ++i)  if(!in[i])
          pq.push(i);
    
        while(!pq.empty()){
          int u = pq.top();  pq.pop();
          mmin = min(mmin, u);
          ans += mmin;
          for(int i = 0; i < G[u].size(); ++i){
            int v = G[u][i];  --in[v];
            if(!in[v])  pq.push(v);
          }
        }
        cout << ans << endl;
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7120202.html
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