problem1 link
对于每一个,找到其在目标串中的位置,判断能不能移动即可。
problem2 link
如果最后的$limit$为$11=(1011)_{2}$,那么可以分别计算值为$(1011)_{2},(1010)_{2},(100x)_{2},(0xxx)_{2}$的答案数,$x$位置表示可以为任意。也就是可以忽略这些位。
当答案固定时,可以用高斯消元求解。
problem3 link
令$d(i,j)$表示点 $i$到点$j$的距离。
使用最小割求解。将每个点拆成$n$个点,第$i$个点拆成$p(i,0),p(i,1),...,p(i,n-1)$.其中$p(i,j)$如果与源点在一个割集,说明$d(0,i)\leq j$为假,与汇点在一个割集说明$d(0,i)\leq j$为真。
所以,如果一个割产生在边$p(i,j-1)\rightarrow p(i,j)$说明最后$d(0,i)=j$.其中边$p(i,j-1)\rightarrow p(i,j)$的代价为$(want[i]-j)^{2}$
有以下边:
(1)对于0点来说,$p(0,0)$与汇点的边流量为无穷大,说明,最后它与汇点在一个割集,所以$d(0,0)\leq 0$为真;
(2)对于$1\leq i < n$号点来说,源点到$p(i,0)$为无穷大(一定不可能),$p(i, n-1)$到汇点为无穷大(一定为真); $p(i,j-1)\rightarrow p(i,j),1\leq j < n)$为$(want[i]-j)^{2}$
(3)如果原来存在边$i\rightarrow j$,那么$p(i,k)\rightarrow p(j,k-1)$为无穷大,表示如果$d(0,i)>k$,那么一定有$d(0,j)>k-1$
code for problem1
#include <string> class FoxAndChess { public: std::string ableToMove(const std::string &begin, const std::string &target) { int n = static_cast<int>(begin.size()); int idx = 0; for (int i = 0; i < n; ++i) { if (begin[i] == ‘L‘ || begin[i] == ‘R‘) { while (idx < n && target[idx] == ‘.‘) { ++idx; } if (idx == n || begin[i] != target[idx] || (begin[i] == ‘L‘ && i < idx) || (begin[i] == ‘R‘ && i > idx)) { return "Impossible"; } ++idx; } } while (idx < n && target[idx] == ‘.‘) { ++idx; } if (idx != n) { return "Impossible"; } return "Possible"; } };
code for problem2
#include <vector> class XorCards { public: long long numberOfWays(const std::vector<long long> &number, long long limit) { const int n = 52; const int m = static_cast<int>(number.size()); long long result = 0; auto GetBit = [](long long t, int b) -> int { return (t & (1ll << b)) != 0 ? 1 : 0; }; { std::vector<std::vector<int>> g(n, std::vector<int>(m + 1, 0)); for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { g[i][j] = GetBit(number[j], i); } g[i][m] = GetBit(limit, i); } result += Gauss(g); } for (int i = 0; i < n; ++i) { if (GetBit(limit, i) == 0) { continue; } std::vector<std::vector<int>> g(n, std::vector<int>(m + 1, 0)); for (int j = i; j < n; ++j) { for (int k = 0; k < m; ++k) { g[j][k] = GetBit(number[k], j); } if (j > i) { g[j][m] = GetBit(limit, j); } } result += Gauss(g); } return result; } private: long long Gauss(std::vector<std::vector<int>> &g) { int n = static_cast<int>(g.size()); int m = static_cast<int>(g[0].size()) - 1; int col = 0; auto HasBit = [&](int start, int col) { for (int i = n - 1; i >= start; --i) { if (g[i][col] != 0) return i; } return -1; }; int row_number = 0; for (int i = 0; i < n && col < m; ++i) { int row = HasBit(i, col); while (row == -1 && col + 1 < m) { row = HasBit(i, ++col); } if (row == -1) { break; } if (row != i) { std::swap(g[i], g[row]); } for (int idx = 0; idx < n; ++idx) { if (idx != i && g[idx][col] != 0) { for (int k = 0; k <= m; ++k) { g[idx][k] ^= g[i][k]; } } } ++col; ++row_number; } for (int i = row_number; i < n; ++i) { if (g[i][m] != 0) { return 0; } } if (m < row_number) { return 0; } return 1ll << (m - row_number); } };
code for problem3
#include <limits> #include <unordered_map> #include <vector> template <typename FlowType> class MaxFlowSolver { static constexpr FlowType kMaxFlow = std::numeric_limits<FlowType>::max(); static constexpr FlowType kZeroFlow = static_cast<FlowType>(0); struct node { int v; int next; FlowType cap; }; public: int VertexNumber() const { return used_index_; } FlowType MaxFlow(int source, int sink) { source = GetIndex(source); sink = GetIndex(sink); int n = VertexNumber(); std::vector<int> pre(n); std::vector<int> cur(n); std::vector<int> num(n); std::vector<int> h(n); for (int i = 0; i < n; ++i) { cur[i] = head_[i]; num[i] = 0; h[i] = 0; } int u = source; FlowType result = 0; while (h[u] < n) { if (u == sink) { FlowType min_cap = kMaxFlow; int v = -1; for (int i = source; i != sink; i = edges_[cur[i]].v) { int k = cur[i]; if (edges_[k].cap < min_cap) { min_cap = edges_[k].cap; v = i; } } result += min_cap; u = v; for (int i = source; i != sink; i = edges_[cur[i]].v) { int k = cur[i]; edges_[k].cap -= min_cap; edges_[k ^ 1].cap += min_cap; } } int index = -1; for (int i = cur[u]; i != -1; i = edges_[i].next) { if (edges_[i].cap > 0 && h[u] == h[edges_[i].v] + 1) { index = i; break; } } if (index != -1) { cur[u] = index; pre[edges_[index].v] = u; u = edges_[index].v; } else { if (--num[h[u]] == 0) { break; } int k = n; cur[u] = head_[u]; for (int i = head_[u]; i != -1; i = edges_[i].next) { if (edges_[i].cap > 0 && h[edges_[i].v] < k) { k = h[edges_[i].v]; } } if (k + 1 < n) { num[k + 1] += 1; } h[u] = k + 1; if (u != source) { u = pre[u]; } } } return result; } MaxFlowSolver() = default; void Clear() { edges_.clear(); head_.clear(); vertex_indexer_.clear(); used_index_ = 0; } void InsertEdge(int from, int to, FlowType cap) { from = GetIndex(from); to = GetIndex(to); AddEdge(from, to, cap); AddEdge(to, from, kZeroFlow); } private: int GetIndex(int idx) { auto iter = vertex_indexer_.find(idx); if (iter != vertex_indexer_.end()) { return iter->second; } int map_idx = used_index_++; head_.push_back(-1); return vertex_indexer_[idx] = map_idx; } void AddEdge(int from, int to, FlowType cap) { node p; p.v = to; p.cap = cap; p.next = head_[from]; head_[from] = static_cast<int>(edges_.size()); edges_.emplace_back(p); } std::vector<node> edges_; std::vector<int> head_; std::unordered_map<int, int> vertex_indexer_; int used_index_ = 0; }; #include <string> class FoxAndCity { public: int minimalCost(const std::vector<std::string> linked, const std::vector<int> want) { constexpr int kMaxCapacity = 1000000; int n = static_cast<int>(linked.size()); MaxFlowSolver<int> solver; auto Index = [&](int u, int k) { return u * n + k; }; int source = -1; int sink = -2; for (int i = 0; i < n; ++i) { if (i == 0) { solver.InsertEdge(Index(i, 0), sink, kMaxCapacity); continue; } solver.InsertEdge(source, Index(i, 0), kMaxCapacity); solver.InsertEdge(Index(i, n - 1), sink, kMaxCapacity); for (int j = 1; j < n; ++j) { solver.InsertEdge(Index(i, j - 1), Index(i, j), (want[i] - j) * (want[i] - j)); } } for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { if (linked[i][j] == ‘Y‘) { for (int k = 1; k < n; ++k) { solver.InsertEdge(Index(i, k), Index(j, k - 1), kMaxCapacity); } } } } return solver.MaxFlow(source, sink); } };
原文地址:https://www.cnblogs.com/jianglangcaijin/p/9384937.html