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题目链接单调队列使用前缀和预处理使用队列维护长度为m的集合从队列中根据单调性在O(1)O(1)O(1)的时间内求得极值#include <iostream>#include <cstring>#include <algorithm>#include <climits>using namespace std;typedef long long LL;const int N = 3e5+6;int q[N];LL s[N];int

题目链接双指针注意：数组初始化特殊值#include <iostream>#include <cstring>#include <algorithm>using namespace std;int n;int cnt[3];char s[200010];int main(){ scanf("%d", &n); while (n -- ){ scanf("%s", s); int

核心：主要是二维数据范围在10的6次方，所以如果直接申请二维空间，会超内存。所以要用一维数组来存，然后转换为二维数组对应的下标。由于C++栈空间默认是1M为了防止暴栈，可以用宽搜#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int...

AcWing题目链接带有路径的最短路径BFS#include <iostream>#include <algorithm>#include <cstdio>#include <vector>#include <queue>using namespace std;typedef pair<int, int> PI...

AcWing题目链接Flood Fill经典题数据较小使用dfs，数据较大使用bfs#include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;const int N = 1005;char g[N][N];in...

AcWing题目链接#include<iostream>#include<algorithm>#include<cstring>using namespace std;const int N=510,INF=0x3f3f3f;int n,m;bool st[N];//状态int g[N][N]; //创建图int d[N];//点之间对距离i...

AcWing题目链接Kruskal算法把所有边按照边权从小到大排序把边权最小的两个顶点使用并查集加入同一个集合#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int M = 2e5+5;int root[M];struct...

Floyd三重循环，一般顶点数不超过200时间复杂度O(n3)O(n^3)O(n3)#include <iostream>#include <cstring>#include <algorithm>using namespace std;int n, m, Q;const int N = 210;const int INF = 1e9;int...

AcWing题目链接#include <iostream>#include <vector>#include <algorithm> #include <cmath>#include <cstring>#include <queue>#include <unordered_map>using nam...

SPFA使用队列来保存更新过的顶点（栈或者优先队列也可以）。只有更新过的顶点才会产生松弛操作。#include <iostream>#include <vector>#include <algorithm> #include <cmath>#include <cstring>#include <queue>#i...

Bellman-Ford求解有负权边的单源最路径问题。时间复杂度O(nm)O(nm)O(nm)通常来说用SPFA注意备份距离数组的使用#include <iostream>#include <cstring>#include <algorithm>using namespace std;int m, n, k;const int N = 5...

AcWing题目链接染色法用染色法判断二分图如果一个图是二分图，当且仅当图中不含奇数环对每个节点进行交错染色，如果染色没有出现矛盾则是二分图。#include <iostream>#include <cstring>#include <algorithm>#include <queue>using namespace std;co...

AcWing题目链接拓扑排序如果用数组模拟队列的话，可以直接输出拓扑序列注意队尾是从-1开始#include <iostream>#include <cstring>using namespace std;const int N = 1e5+5;int h[N], e[N], ne[N];int q[N], d[N]={0};int idx = 0;i...

AcWing题目链接BFS模板题用数组模拟队列#include <iostream>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int N = 1e5+5;const int M = 2e5+5;int h[N], ...

注意ne数组和e数组的大小是顶点数的两倍每个顶点遍历一遍，搜索中计算和返回。#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 1e5+5, M = N*2;bool st[N]={false}; int h[N],...

AcWing题目链接BFS有个小剪枝，对于BFS第一次到达的点才操作，不然就终止。#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <queue>using namespace std;typedef pair&l...

AcWing题目链接DFS#include <iostream>#include <cstdio>#include <cmath>using namespace std;const int N = 20;int n;int pos[N];void dfs(int u){ if(u == n){ //8个皇后都放好了 放在前u行 for(...

AcWing题目链接Trie树模板2#include <iostream>#include <string>#include <vector>#include <cctype>#include <cstdio>#include <cstring>#include <algorithm>#inclu...

AcWing题目链接#include <iostream>using namespace std;typedef long long LL;const int N = 1e6+5;int primes[N];bool st[N];int phi[N];int k = 0;LL get_eulers(int n){ phi[1] = 1; for(int...

AcWing题目链接埃氏筛法时间复杂度O(nloglogn)O(nloglogn)O(nloglogn)#include <iostream>using namespace std;const int N = 1e6+5;int primes[N];bool st[N];int k = 0;void get_primes(int n){ for(int i =...

AcWing题目链接欧拉函数ϕ(n)\phi(n)ϕ(n)表示1∼n1\sim n1∼n中与nnn互质的个数n=p1α1p2α2…pkαkn = p_1^{\alpha_1}p_2^{\alpha_2} \dots p_k^{\alpha_k}n=p1α1​​p2α2​​…pkαk​​ϕ(n)=n(1−1p1)(1−1p2)…(1−1pk)\phi(n)= n(1-\frac{1}{p_1...

//a[i]+a[j]+i-j = (a[i] + i) + (a[j] - j)#include <iostream>#include <algorithm>using namespace std;const int N = 1e5+7;int arr[N];int main(){ int n; cin>>n; for(in...

状态机注意买卖的定义#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 100010, M = 110, INF = 0x3f3f3f3f;int n, m;int w[N];int f[N][M][2];i...

自动机0——> 00——>11——>0因为不许抢两个相邻商店 所以不允许从1到1#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 100010, INF = 0x3f3f3f3f;int n...

xy zx位置上的最大值取决于y和z的最大值加上x本身的值，可以用一维数组从底至顶迭代。边界条件是最底层的最值就是这一行本身。#include <iostream>#include <algorithm>using namespace std;const int N = 505;int a[N][N], g[N];int main(){ int ...

题目链接class Solution {public: int Fibonacci(int n) { int f[40]; f[0] = 0; f[1] = 1; for(int i = 2; i <= n; i++){ f[i] = f[i-1] + f[i-2]; ...

题目链接class MyQueue {public: /** Initialize your data structure here. */ MyQueue() { } stack<int> stk, cache; /** Push element x to the back of queue. */ void co...

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode *father; * TreeNode(int x) : val(x), left(NULL), righ...

题目链接/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */...

题目链接/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: vector&...

#include <iostream>#include <string>#include <cctype>using namespace std;int u;string str;string dfs(){ string res; while(u < str.size()){ char c = str[u]; ...

题目链接如：1234 / 7商：0176 余2去掉前面的0 但是也要至少保留1位 跟减法类似#include <iostream>#include <string>#include <algorithm>using namespace std;int main(){ string a,res; int b; cin>...

题目链接快速排序三个步骤1.选分界点2.确定左右两段3.递归处理左右两段#include <iostream>using namespace std;const int maxn = 1e5+5;int arr[maxn];void quickSort(int l, int r){ if(l >= r) return; int x = arr[l+r&gt...

二分模板题题目链接两段性#include <string>#include <vector>#include <iostream>#include <algorithm> using namespace std;vector<int> nums;int main(){ int n, k, x; cin>&gt...

#include<iostream>using namespace std;typedef unsigned long long ULL;int main(){ ULL a, b, p; cin >> a >> b >> p; ULL res = 0; while(b){ if(b&am...

#include <iostream>#include <vector>#include <cstring>using namespace std;vector<int> res, nums;bool vis[10];void dfs(int k, int u) { if (u == nums.size()) { res.push_...

AcWing题目链接匈牙利算法模板题最大匹配：边数最多的匹配完美匹配：所有顶点都匹配上了。完美匹配一定是最大匹配。时间复杂度O(mn)O(mn)O(mn)，实际运行效率远小于这个值。#include <iostream>#include <string>#include <vector>#include <cstring>#inc...

题目链接#include <iostream>#include <vector>#include <algorithm> #include <cmath>#include <cstring>using namespace std;int main(){ int n, a, b, p; cin>>n; whi...

题目链接朴素Dijkstra算法有自环和重边#include <iostream>#include <vector>#include <algorithm> #include <cmath>#include <cstring>using namespace std;const int N = 505;const int...

AcWing题目链接模板题#include <iostream>#include <vector>#include <algorithm> #include <cmath>#include <cstring>using namespace std;const int N = 1e5+7;int nums[N];int ...