Handsomk

题目链接每加入一个条件就要进行一次拓扑排序有以下三种情况：条件不足：多个入度为0、多个出度为0、当前能得出的序列长度小于总字母数量出现环路成功判断测试点14 6C<DC<BB<AC<DD<AA<AInconsistency found after 6 relations.#include <iostream>#include <queue>#include <vector>#includ

题目链接堆优化迪杰斯特拉#include <iostream>#include <queue>#include <vector>#include <cstring>using namespace std;typedef pair<int, int> pii;const int N = 1010;struct node { int to, d, m;};vector<vector<node>&gt

题目链接要先计算连通分支的个数，最终只有一个连通分支即可#include <iostream>#include <algorithm>using namespace std;const int N = 5010;struct node { int wei, x, y, sta; bool operator < (const node& t) const { return wei < t.wei; }}e[

题目链接最小生成树模板题#include <iostream>#include <algorithm>using namespace std;const int N = 5010;struct node { int wei, x, y; bool operator < (const node& t) const { return wei < t.wei; }}e[N];int f[110];int

简单的并查集，主要复习模板题目链接#include <iostream>using namespace std;const int N = 1010;int f[N];int find(int x) { if (x != f[x]) return f[x] = find(f[x]); return f[x];}void merge(int a, int b) { int ta = find(a); int tb = find(b); .

题目链接#include <iostream>using namespace std;struct node { int val; node *l, *r; node(int x) : val(x), l(nullptr), r(nullptr) {}};int insertTree(node *&root, int x) { if (root == nullptr) { root = new node(x);

题目链接#include <iostream>using namespace std;struct node { char c; node *l, *r; node(char x ) : c(x), l(nullptr), r(nullptr) {}};node* buildTree(string pre, string mid, int pl, int pr, int ml, int mr) { if (pl > pr) {

题目链接#include <iostream>#include <vector>using namespace std;struct node { char c; node *l; node *r; node(char x) : c(x), l(nullptr), r(nullptr) {}};node* buildTree(int &pos, string str) { if (pos >= str.size()

题目链接关键点：注意输出时的字母要按字典序，具体就是深搜时的方向排序问题#include <iostream>#include <queue>#include <algorithm>#include <cstring>#include <vector>using namespace std;struct node { int x, y; node(int a, int b) : x(a), y(b){};}

清华上机题链接#include <iostream>#include <string>#include <queue>#include <unordered_map>using namespace std;struct node { string s; int cnt; node(string x, int y) : s(x), cnt(y) {};};unordered_map<string, bool>

题目链接#include <iostream>#include <climits>#include <queue>using namespace std;const int N = 100010;struct node { int pos, t; node(int x, int y) : pos(x), t(y){}};int ans;bool flag[N];queue<node> q;void bfs(int cu

时隔一年半重写，区间贪心按右端点排序题目链接#include <iostream>#include <vector>#include <algorithm>using namespace std;struct node { int st, ed; bool operator < (const node &t) const { if (ed != t.ed) return ed < t.ed; .

题目-矩阵幂模板#include <iostream>#include <algorithm>#include <string>#include <cstring>using namespace std;const int N = 15;struct Matrix { int v[N][N]; int row, col;};Matrix mutil(Matrix a, Matrix b) { Matrix c;

约数的个数约数定理#include <iostream>using namespace std;const int N = 1e5 + 10;int primes[N];bool st[N];int cnt = 0;void getPrimes() { for (int i = 2; i < N; i++) { if (!st[i]) primes[++cnt] = i; for (int j = 1; i * prime

关键点: 每个合数只被最小的质因数筛掉int primes[N];bool st[N];int cnt = 0;void getPrimes() { for (int i = 2; i < N; i++) { if (!st[i]) primes[++cnt] = i; for (int j = 1; i * primes[j] < N; j++) { //i的质因数肯定比primes[j]大，所以就保证了primes.

10进制 VS 2进制一系列高精度操作#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;//高精度除以低精度（2）void div(string &s) { string ans = ""; int r = 0; for (int i = 0; i < s.leng

进制转换清华大学上机题十进制转二进制，十进制最多有30位主要用高精度除以低精度#include <iostream>#include <vector>#include <algorithm>using namespace std;void div(vector<int> &v) { vector<int> c; int r = 0; for (int i = 0; i < v.size(

KMP算法