topcoder srm 615 div1

problem1 linknode

對於數字$x$,檢驗每一個知足$x=y*2^{t}$的$y$可否變成$x$便可。code

problem2 linkblog

若是起點到終點有一條長度爲$L$的路徑,那麼就存在長度爲$L+kR$的路徑。其中$R$爲從路徑上某點轉一圈再回到這一點的環的長度。ip

爲了保證老是存在這個環,能夠令這個環爲從起點出發再回到起點。因此若是有一條長度爲$d$的邊$0\rightarrow t$,那麼能夠令$R=2d$,即$0\rightarrow t \rightarrow 0$.get

只須要記錄起點到達某個點長度模$R$的最短路便可。即用$f[m][u]$表示從0到$u$的最小的知足$m+kR$的路徑長度。string

只要$f[T$%$R][N-1] \leq T$便可。it

problem3 linkast

紅色和綠色須要配對出現。因此能夠將一個紅色一個綠色看做一個總體。那麼就是$M$組中每組要出現$D$個配對的總體。class

從前向後進行動態規劃,只統計出現了多少個配對的總體以及還有多少個只配對了一半的。統計

這裏有個問題是每一組中的$D$個總體不能交叉出現。爲了作到這一點,只須要配對了一半的個數不超過$M$便可.

 

code for problem1

#include <set>
#include <vector>

class AmebaDiv1 {
 public:
  int count(const std::vector<int> &X) {
    auto Get = [&](int t) {
      for (auto e : X) {
        if (e == t) {
          t *= 2;
        }
      }
      return t;
    };

    std::set<int> all(X.begin(), X.end());
    int result = 0;
    for (auto e : all) {
      bool tag = (Get(1) != e);
      int t = e;
      while (t > 1) {
        if (Get(t) == e) {
          tag = false;
          break;
        }
        t /= 2;
      }
      if (tag) {
        ++result;
      }
    }

    return result;
  }
};

code for problem2

#include <cstring>
#include <set>
#include <string>
#include <vector>

constexpr int MAXD = 20000;
constexpr int MAXN = 50;

long long dist[MAXD][MAXN];

class LongLongTripDiv1 {
 public:
  std::string isAble(int N, const std::vector<int> &A,
                     const std::vector<int> &B, const std::vector<int> &D,
                     long long T) {
    int n = static_cast<int>(A.size());
    int d = MAXD + 1;
    for (int i = 0; i < n; ++i) {
      if (A[i] == 0 || B[i] == 0) {
        d = std::min(d, D[i]);
      }
    }

    if (d == MAXD + 1) {
      return "Impossible";
    }

    std::set<std::pair<long long, std::pair<int, int>>> que;
    memset(dist, -1, sizeof(dist));
    auto Insert = [&](long long dis, int u, int v) {
      auto iter = que.find({dist[u][v], {u, v}});
      if (iter != que.end()) {
        que.erase(iter);
      }
      que.insert({dis, {u, v}});
      dist[u][v] = dis;
    };
    Insert(0, 0, 0);
    while (!que.empty()) {
      auto node = que.begin()->second;
      que.erase(que.begin());
      int u = node.second;
      for (int i = 0; i < n; ++i) {
        if (A[i] != u && B[i] != u) {
          continue;
        }
        int v = A[i] + B[i] - u;
        int t = (node.first + D[i]) % (2 * d);
        long long new_dist = dist[node.first][u] + D[i];
        if (dist[t][v] == -1 || new_dist < dist[t][v]) {
          Insert(new_dist, t, v);
        }
      }
    }
    auto min_dist = dist[T % (d * 2)][N - 1];
    return (min_dist != -1 && min_dist <= T) ? "Possible" : "Impossible";
  }
};

code for problem3

#include <string>

constexpr int MOD = 1000000007;

int f[5005][50][51];

class AlternativePiles {
 public:
  int count(const std::string &C, int M) {
    if (Red(C[0])) {
      f[0][0][1] += 1;
    }
    if (Blud(C[0])) {
      f[0][0][0] += 1;
    }
    for (size_t idx = 1; idx < C.size(); ++idx) {
      char c = C[idx];
      for (int x = 0; x < M; ++x) {
        for (int y = 0; y <= M; ++y) {
          int t = f[idx - 1][x][y];
          if (t == 0) {
            continue;
          }
          if (Red(c) && y + 1 <= M) {
            (f[idx][x][y + 1] += t) %= MOD;
          }
          if (Blud(c)) {
            (f[idx][x][y] += t) %= MOD;
          }
          if (Green(c) && y > 0) {
            (f[idx][(x + 1) % M][y - 1] += t) %= MOD;
          }
        }
      }
    }
    return f[C.size() - 1][0][0];
  }

 private:
  bool Red(char c) { return c == 'R' || c == 'W'; }

  bool Green(char c) { return c == 'G' || c == 'W'; }

  bool Blud(char c) { return c == 'B' || c == 'W'; }
};
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