AtCoder Beginner Contest 399 (A - F) A - Hamming Distance#include <bits/stdc++.h> using namespace std; using LL = long long; void solve() { int n; string s, t; cin >> n >> s >> t; int ans =… Raven...发表于AtCod... AtCoder Beginner Conte...
void solve() { int n; std::cin >> n; std::vector<std::string> s(n); for (int i = 0; i < n; ++ i) { std::cin >> s[i]; } std::vector<std::array<std::vector<int>, 26>> adj1(n), adj2(n); for (int i = 0; i < n; ++ i) { for (int j = 0; j <...
从起点u和终点v分别出发,若路径上有奇数个点,则最终汇合在一个点上,即(i,i);若有偶数个点,则汇合在一对有边连接的点上,即i,j且i,j之间有边。则我们可以对每个(i,i)和(i,j)分别向(u,v)扩展,取最短路径。详情见代码 代码# 点击查看代码 #include<bits/stdc++.h>#include<unordered_set>#include<...
KAJIMA CORPORATION CONTEST 2024(AtCoder Beginner Contest 340) A - Arithmetic Progression 代码: #include <bits/stdc++.h> using namespace std; using