Love Rescue 题意:Valya 和 Tolya 是一对情侣, 他们的T恤和头巾上都有小写字母,但是女朋友嫌弃男朋友上T恤上的字不和她的头巾上的字一样,就很生气, 然后来了一个魔法师, 它可以购买符文, 每个符文可以让一个字母变成另外一个字母,魔法师想购买足够的符文使得他们的字母串相同, 因为买符文要钱, 现在需要求最小的购买符文数目,和符文形式, 多种答案可以输出任意一种. 题解:用并查集将有联系的字母分到一组, 然后将一组内的符文进行串联,这样每一组内的任意2个符文, 都可以由上面的…

E. Maximize! time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given a multiset S consisting of positive integers (initially empty). There are two kind of queries: Add a positive int…

A #include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!!\n") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vect…

D. Chip 'n Dale Rescue Rangers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/problem/D Description A team of furry rescue rangers was sitting idle in their hollow tree when suddenly they received a signal of distress.…

传送门:http://codeforces.com/contest/939/problem/D 本题是一个数据结构问题——并查集(Disjoint Set). 给出两个长度相同,且仅由小写字母组成的字符串S=s[1..n].T=t[1..n].已知一个无序对(u,v)可以完成任意次的以下转换操作:u→v,v→u.求将字符串S转换为T所需要的最少的无序对的数目,并打印出相应可行的方案下的所有无序对. 首先构造一张无向图G=<V,E>,表示S→T的状态转换图:结点集V={‘a’,’b’,’c’,.…

题目链接: B. Chip 'n Dale Rescue Rangers time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A team of furry rescue rangers was sitting idle in their hollow tree when suddenly they received a signa…

题目链接: 题目 B. Chip 'n Dale Rescue Rangers time limit per test:1 second memory limit per test:256 megabytes 问题描述 A team of furry rescue rangers was sitting idle in their hollow tree when suddenly they received a signal of distress. In a few moments they…

D. Love Rescue time limit per test2 seconds memory limit per test256 megabytes Problem Description Valya and Tolya are an ideal pair, but they quarrel sometimes. Recently, Valya took offense at her boyfriend because he came to her in t-shirt with let…

题目链接:https://codeforces.com/contest/1420/problem/D 前言 之前写过这场比赛的题解,不过感觉这一题还可以再单独拿出来好好捋一下思路. 题意 给出 $n$ 个闭区间,问 $k$ 个区间共区间共有多少种情况. 题解一 以区间为单位进行考虑,排序+优先队列. 将所有区间以左端点为第一关键字,右端点为第二关键字从小到大排序,优先队列中存储不小于当前区间左端点的之前区间的右端点,每个区间对答案的贡献即 $C_{(pque.size(),\ k - 1)}$…

这题可以o(1)推出公式,也可以二分答案+验证. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> using namespace std; double X1,X2,Y1,Y2; double v,t; double vx,vy,wx,wy; bool check(double mid) { double tmpx,…