活动安排问题

题目描述

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解题代码

vector<bool> greedySelector(vector<vector<int>>& intervals) {int n = intervals.size();// 将活动区间按结束时间的从小到大排序auto cmp = [](vector<int>& interval1, vector<int>& interval2) {return interval1[1] < interval2[1];};sort(intervals.begin(), intervals.end(), cmp);vector<bool> res(n, false);// 结束时间最早的活动必定位于某个最优解之中int minStart = intervals[0][1];res[0] = true;for (int i = 1; i < n; ++i) {if (intervals[i][0] >= minStart) { // 将不重叠的活动加入最优解集res[i] = true;minStart = intervals[i][1];}}return res;
}

最优装载

题目描述

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解题代码

vector<bool> optimisedLoading(vector<int>& weight, int c) {int n = weight.size();vector<bool> select(n, false);// 定义一个小顶优先队列，使得对于i，若其weight[i]最小，则排在队列的队头priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;for (int i = 0; i < n; ++i) {// 构建二元组<重量,下标>并放入优先队列q.emplace(weight[i], i);}for (int i = 0; i < n; ++i) {auto [w, idx] = q.top(); // （C++17语法）取队头元素的w和对应下标q.pop();if (c < w) break; // 无法继续装载select[idx] = true; // 选择装载该货物c -= w; // 剩余载货量减少}return select;
}

哈夫曼编码

题目描述

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解题代码

struct HuffmanNode {int left, right; // 左右结点int parent; // 父结点int weight; // 权重char data; // 数据HuffmanNode(int left = -1, int right = -1, int parent = -1, int weight = 0, char data = '*'): left(left), right(right), parent(parent), weight(weight), data(data) {}
};vector<HuffmanNode> createHuffmanTree(vector<int>& weight, vector<char>& data) {int n = weight.size();vector<HuffmanNode> huffmanTree(2 * n);// 定义一个小顶优先队列，使得对于i，若其weight[i]最小，则排在队列的队头priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;for (int i = 0; i < n; ++i) { // 初始化哈夫曼树和优先队列huffmanTree[i].data = data[i];huffmanTree[i].weight = weight[i];q.emplace(weight[i], i);}for (int i = 0; i < n - 1; ++i) {auto [weight1, idx1] = q.top(); // 取权值最小结点q.pop();auto [weight2, idx2] = q.top(); // 取权值第二小结点q.pop();// 创建两结点的父结点，其下标为n+ihuffmanTree[idx1].parent = n + i;huffmanTree[idx2].parent = n + i;// 初始化该父结点的相关信息huffmanTree[n + i].left = idx1;huffmanTree[n + i].right = idx2;huffmanTree[n + i].weight = weight1 + weight2;// 将该父结点的<权值,下标>加入优先队列，以便进行贪心选择q.emplace(weight1 + weight2, n + i);}return huffmanTree;
}void printHuffmanCode(vector<HuffmanNode>& huffmanTree) {stack<int> s;for (int i = 0; i < huffmanTree.size() / 2; ++i) {int cur = i; // 当前结点下标int pre = huffmanTree[cur].parent; // 当前结点的父结点的下标while (pre != -1) {if (huffmanTree[pre].left == cur) {s.emplace(0); // 当前结点为其父结点的左孩子}else {s.emplace(1); // 当前结点为其父结点的右孩子}// 轮换下标cur = pre;pre = huffmanTree[cur].parent;}// 打印相应的哈夫曼编码cout << huffmanTree[i].data << " ";while (!s.empty()) {cout << s.top();s.pop();}cout << endl;}
}

单源最短路径

题目描述

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解题代码

图的定义

struct MGraph {vector<char> vertices; // 顶点数组vector<vector<int>> edges; // 邻接矩阵
};

BellmanFord

此算法可适用于含有负权值边的图。

// G:图	start:源点	dist:最短路径
bool BellmanFord(MGraph& G, int start, vector<int>& dist) {int n = G.vertices.size();// 初始化最短路径dist.assign(n, INT32_MAX);for (int j = 0; j < n; ++j) {dist[j] = G.edges[start][j];}for (int t = 0; t < n - 1; ++t) { // 松弛次数tfor (int i = 0; i < n; ++i) { // 边的起点ifor (int j = 0; j < n; ++j) { // 边的终点jif (G.edges[i][j] != INT32_MAX && dist[i] != INT32_MAX&& dist[i] + G.edges[i][j] < dist[j]) {dist[j] = dist[i] + G.edges[i][j]; // 松弛操作}}}}for (int i = 0; i < n; ++i) {for (int j = 0; j < n; ++j) {// 若执行完算法后仍然存在非最短路径，则该图存在权值为负的环路，无最短路径if (G.edges[i][j] != INT32_MAX && dist[i] != INT32_MAX&& dist[i] + G.edges[i][j] < dist[j]) {return false;}}}return true;
}

Dijkstra

本算法仅适用于所有边的权值均为正的图。

// G:图	start:源点	dist:最短路径
void Dijkstra(MGraph& G, int start, vector<int>& dist) {int n = G.vertices.size();// 初始化最短路径dist.assign(n, INT32_MAX);vector<int> pre(n, -1); // 前驱数组vector<bool> visited(n, false); // 访问集，表示对应顶点最短路径是否已经找到visited[start] = true;// 小顶优先队列，元素为<dist[j],j>priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;// 初始化最短路径for (int j = 0; j < n; ++j) {dist[j] = G.edges[start][j];q.emplace(dist[j], j);}while (!q.empty()) {int i = q.top().second; // 贪心选择最近结点iq.pop();visited[i] = true; // 结点i最短路径已得到for (int j = 0; j < n; ++j) { // 利用结点i进行松弛操作if (visited[j]) continue; // 结点j已得到最短路径，无需松弛if (G.edges[i][j] != INT32_MAX && dist[i] != INT32_MAX&& dist[i] + G.edges[i][j] < dist[j]) {dist[j] = dist[i] + G.edges[i][j]; // 松弛操作pre[j] = i; // 更新前驱结点q.emplace(dist[j], j); // 加入优先队列}}}// 打印源点到各结点的最短路径stack<int> s;for (int i = 0; i < n; ++i) {if (i == start) continue;if (dist[i] == INT32_MAX) {cout << "inf" << ": " << G.vertices[start] << "->" << G.vertices[i] << endl;}else {cout << dist[i] << ": " << G.vertices[start] << "->";int x = pre[i];while (x != -1) {s.emplace(x);x = pre[x];}while (!s.empty()) {cout << G.vertices[s.top()] << "->";s.pop();}cout << G.vertices[i] << endl;}}
}

最小生成树

题目描述

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解题代码

Kruskal

void Kruskal(MGraph& G) {int n = G.vertices.size();// 边集，元素为<权值weight，起点u，终点v>vector<tuple<int, int, int>> edges;for (int i = 0; i < n; ++i) {for (int j = i + 1; j < n; ++j) {if (G.edges[i][j] != INT32_MAX) {// 将边加入边集edges.emplace_back(G.edges[i][j], i, j);}}}// 对边集按权值大小进行升序排序sort(edges.begin(), edges.end());// 简单并查集，father[x]存放x的父结点vector<int> father(n);// 寻找x所在集合的父结点（所在连通分量编号）auto findFather = [&](int x) {int f = father[x];while (f != x) {x = f;f = father[x];}return f;};for (int i = 0; i < n; ++i) {father[i] = i; // 初始父结点为自身}int cnt = 0; // 已找到的边个数for (int i = 0; cnt <= n - 1 && i < edges.size(); ++i) {auto [weight, u, v] = edges[i];int fu = findFather(u);int fv = findFather(v);// 若u和v父结点相同（即u和v位于一个连通分量中），若选择加入边uv，则会导致回路if (fu != fv) { cout << G.vertices[u] << " - " << G.vertices[v] << " : " << weight << endl;father[fu] = fv; // 两个连通分量合并为一个++cnt;}}
}

Prim

void Prim(MGraph& G) {int n = G.vertices.size();// minDist[i]表示结点i距离MST最近距离vector<int> minDist(n, INT32_MAX);// connected[i]表示在MST中与结点i相连的结点vector<int> connected(n, 0);// 表示结点i是否已加入MSTvector<bool> visited(n, false);visited[0] = true;for (int j = 1; j < n; ++j) {// 初始化最近距离minDist[j] = G.edges[0][j];}for (int i = 1; i < n; ++i) {// 寻找距离MST的最近结点kint minVal = INT32_MAX, k = -1;for (int j = 1; j < n; ++j) {if (!visited[j] && minDist[j] < minVal) {minVal = minDist[j];k = j;}}if (k == -1) break;// 将结点k加入MST中visited[k] = true;cout << G.vertices[connected[k]] << " - " << G.vertices[k] << " : " << minVal << endl;// 更新minDist数组和connected数组for (int j = 1; j < n; ++j) {if (!visited[j] && G.edges[k][j] < minDist[j]) {minDist[j] = G.edges[k][j];connected[j] = k;}}}
}