• #6510. 「雅礼集训 2018 Day8」A [最小树形图]


    建一个超级源点,建新边,然后就可以对于这个超级源点跑最小树形图了

    // powered by c++11
    // by Isaunoya
    #include <bits/stdc++.h>
    #define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
    #define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
    using namespace std;
    using db = double;
    using ll = long long;
    using uint = unsigned int;
    #define Tp template
    using pii = pair<int, int>;
    #define fir first
    #define sec second
    Tp<class T> void cmax(T& x, const T& y) {
      if (x < y) x = y;
    }
    Tp<class T> void cmin(T& x, const T& y) {
      if (x > y) x = y;
    }
    #define all(v) v.begin(), v.end()
    #define sz(v) ((int)v.size())
    #define pb emplace_back
    Tp<class T> void sort(vector<T>& v) { sort(all(v)); }
    Tp<class T> void reverse(vector<T>& v) { reverse(all(v)); }
    Tp<class T> void unique(vector<T>& v) { sort(all(v)), v.erase(unique(all(v)), v.end()); }
    const int SZ = 1 << 23 | 233;
    struct FILEIN {
      char qwq[SZ], *S = qwq, *T = qwq, ch;
    #ifdef __WIN64
    #define GETC getchar
    #else
      char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
    #endif
      FILEIN& operator>>(char& c) {
        while (isspace(c = GETC()))
          ;
        return *this;
      }
      FILEIN& operator>>(string& s) {
        while (isspace(ch = GETC()))
          ;
        s = ch;
        while (!isspace(ch = GETC())) s += ch;
        return *this;
      }
      Tp<class T> void read(T& x) {
        bool sign = 0;
        while ((ch = GETC()) < 48) sign ^= (ch == 45);
        x = (ch ^ 48);
        while ((ch = GETC()) > 47) x = (x << 1) + (x << 3) + (ch ^ 48);
        x = sign ? -x : x;
      }
      FILEIN& operator>>(int& x) { return read(x), *this; }
      FILEIN& operator>>(ll& x) { return read(x), *this; }
    } in;
    struct FILEOUT {
      const static int LIMIT = 1 << 22;
      char quq[SZ], ST[233];
      int sz, O;
      ~FILEOUT() { flush(); }
      void flush() {
        fwrite(quq, 1, O, stdout);
        fflush(stdout);
        O = 0;
      }
      FILEOUT& operator<<(char c) { return quq[O++] = c, *this; }
      FILEOUT& operator<<(string str) {
        if (O > LIMIT) flush();
        for (char c : str) quq[O++] = c;
        return *this;
      }
      Tp<class T> void write(T x) {
        if (O > LIMIT) flush();
        if (x < 0) {
          quq[O++] = 45;
          x = -x;
        }
        do {
          ST[++sz] = x % 10 ^ 48;
          x /= 10;
        } while (x);
        while (sz) quq[O++] = ST[sz--];
      }
      FILEOUT& operator<<(int x) { return write(x), *this; }
      FILEOUT& operator<<(ll x) { return write(x), *this; }
    } out;
    #define int long long
    
    int n, m;
    const int maxn = 5e2 + 25;
    const int maxm = maxn * maxn;
    int u[maxm], v[maxm], w[maxm];
    int from[maxn], mnin[maxn], col[maxn];
    bool vis[maxn], ins[maxn];
    int nn, mm;
    void dfs(int x) {
      if (!x || !from[x] == -1) return;
      vis[x] = ins[x] = 1;
      if (!vis[from[x]])
        dfs(from[x]);
      else if (ins[from[x]]) {
        col[x] = ++nn;
        for (int i = from[x]; i != x; i = from[i]) col[i] = nn;
      }
      ins[x] = 0;
      if (!col[x]) col[x] = ++nn;
    }
    
    const int inf = 1e15;
    signed main() {
      // code begin.
      in >> n >> m;
      rep(i, 1, m) in >> u[i] >> v[i] >> w[i];
      rep(i, 1, n) { u[++m] = 0, v[m] = i, w[m] = inf; }
      int ans = 0;
      while (1) {
        rep(i, 1, n) mnin[i] = inf << 1, from[i] = -1;
        rep(i, 1, m) if (w[i] < mnin[v[i]]) { mnin[v[i]] = w[i], from[v[i]] = u[i]; }
        rep(i, 0, n) vis[i] = col[i] = 0;
        nn = mm = 0;
        rep(i, 1, n) {
          ans += mnin[i];
          if (!vis[i]) dfs(i);
        }
        if (n == nn) break;
        rep(i, 1, m) {
          if (col[u[i]] != col[v[i]]) {
            w[++mm] = w[i] - mnin[v[i]];
            u[mm] = col[u[i]], v[mm] = col[v[i]];
          }
        }
        n = nn, m = mm;
      }
      if (ans >= 2 * inf)
        out << -1 << '
    ';
      else
        out << ans - inf << '
    ';
      return 0;
      // code end.
    }
    
  • 相关阅读:
    如何解决无法成功git commit 和git push
    orleans 项目调试注意
    silo 主机 配置
    asp.net core 项目引用包版本问题
    C# async 方法怎么被正确的消费 (新篇)
    C# 虚方法 复习
    C# dynamic 适用场景进一步说明
    [MySQL]
    C# Subject 观察者模式
    C# 协变与逆变
  • 原文地址:https://www.cnblogs.com/Isaunoya/p/12390846.html
Copyright © 2020-2023  润新知