Codeforces Round #361 (Div. 2) B bfs处理】的更多相关文章

B. Mike and Shortcuts time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Recently, Mike was very busy with studying for exams and contests. Now he is going to chill a bit by doing some sight…
B. Mike and Shortcuts 题目连接: http://www.codeforces.com/contest/689/problem/B Description Recently, Mike was very busy with studying for exams and contests. Now he is going to chill a bit by doing some sight seeing in the city. City consists of n inter…
B. Mike and Shortcuts time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Recently, Mike was very busy with studying for exams and contests. Now he is going to chill a bit by doing some sight…
B - Mike and Shortcuts Description Recently, Mike was very busy with studying for exams and contests. Now he is going to chill a bit by doing some sight seeing in the city. City consists of n intersections numbered from 1 to n. Mike starts walking fr…
D. Phillip and Trains time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The mobile application store has a new game called "Subway Roller". The protagonist of the game Philip is located…
A 脑筋急转弯 // #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <cstdio> #include <cstring> #include <sstream> #include <string> #include <algorithm> #include <list> #incl…
题目连接:http://codeforces.com/contest/688/problem/C 题意:给你一些边,问你能否构成一个二分图 题解:二分图:二分图又称作二部图,是图论中的一种特殊模型. 设G=(V,E)是一个无向图,如果顶点V可分割为两个互不相交的子集(A,B),并且图中的每条边(i,j)所关联的两个顶点i和j分别属于这两个不同的顶点集(i in A,j in B),则称图G为一个二分图. 直接上一个DFS就搞定了 #include<cstdio> #include<set…
E. Mike and Geometry Problem 题目连接: http://www.codeforces.com/contest/689/problem/E Description Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geometry problem. Let's define f([l, r]) = r - l + 1 to…
D. Friends and Subsequences 题目连接: http://www.codeforces.com/contest/689/problem/D Description Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on their o…
C. Mike and Chocolate Thieves 题目连接: http://www.codeforces.com/contest/689/problem/C Description Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible! Aside from loving sweet things, thieves from t…
A. Mike and Cellphone 题目连接: http://www.codeforces.com/contest/689/problem/A Description While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was no worry as he bought a cheap replacement phone with an old-fas…
任意门:http://codeforces.com/contest/689/problem/E E. Mike and Geometry Problem time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Mike wants to prepare for IMO but he doesn't know geometry, so…
D - Friends and Subsequences Description Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on their own, but together (with you) — who knows? Every one of…
C - Mike and Chocolate Thieves Description Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible! Aside from loving sweet things, thieves from this area are known to be very greedy. So after a thie…
A - Mike and Cellphone Description While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was no worry as he bought a cheap replacement phone with an old-fashioned keyboard. The keyboard has only ten digital eq…
C 给出一个m 此时有 四个数 分别为x k*x k*k*x k*k*k*x k大于1 x大于等于1 要求求出来一个最小的值n 使其满足 这四个数中的最大值小于n 这四个数可能的组数为m 可以看出这四个数递增 所以最大值必定是第四个 所以我们二分n 需要注意的是longlong可以定义到1e18 初始应当是l小于等于可能的最小值(0) r大于等于可能的最大值(1e18) 如果初始范围就定义错误的话 二分不会出现正确答案 在这里由于控制l=mid+1或者r=mid-1 这一类的时候 容易丢失mid…
A - Mike and Cellphone 问有没有多解,每个点按照给出的序列用向量法跑一遍 #include<cstdio> #include<cstring> #include<queue> #include<cstdlib> #include<algorithm> #include<vector> #include<cmath> #include<map> using namespace std; ty…
题目大意:给你两个长度为n的数组a, b,问你有多少个问你有多少个区间满足 a中最大值等于b中最小值. 思路:我本来的想法是用单调栈求出每个点的管辖区间,然后问题就变成了巨麻烦的线段覆盖问题,就爆炸写了 一晚上假算法.正解就是枚举一个端点,然后二分找右端点的区间,因为满足一个很神奇的单调性,然后st表维护 一下区间最值就好了. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #def…
E. Mike and Geometry Problem time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geometry pro…
题目链接:传送门 题目大意:给你n个区间,求任意k个区间交所包含点的数目之和. 题目思路:将n个区间都离散化掉,然后对于一个覆盖的区间,如果覆盖数cnt>=k,则数目应该加上 区间长度*(cnt与k的组合数) ans=ans+(len*C(cnt,k))%mod; #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorith…
A. Mike and Cellphone time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was no worry as he bough…
题目传送门 /* 题意:告诉起点终点,踩一次, '.'变成'X',再踩一次,冰块破碎,问是否能使终点冰破碎 DFS:如题解所说,分三种情况:1. 如果两点重合,只要往外走一步再走回来就行了:2. 若两点相邻, 那么要不就是踩一脚就破了或者踩一脚走开再走回来踩一脚破了:3. 普通的搜索看是否能到达, 若能还是要讨论终点踩几脚的问题:) DFS 耗时大,险些超时,可以用BFS来做 */ #include <cstdio> #include <algorithm> #include &l…
题目传送门 /* 题意:问最少替换'*'为'.',使得'.'连通的都是矩形 BFS:搜索想法很奇妙,先把'.'的入队,然后对于每个'.'八个方向寻找 在2*2的方格里,若只有一个是'*',那么它一定要被替换掉 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <queue> using namespace std; ;…
Codeforces Round #599 (Div. 2) D. 0-1 MST Description Ujan has a lot of useless stuff in his drawers, a considerable part of which are his math notebooks: it is time to sort them out. This time he found an old dusty graph theory notebook with a descr…
Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…
Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/output 1 s, 256 MB  x2377 B Queue standard input/output 2 s, 256 MB  x1250 C Hacking Cypher standard input/output 1 s, 256 MB  x740 D Chocolate standard in…
Codeforces Round #117 (Div. 2) 代码 Codeforces Round #117 (Div. 2) A. Battlefield any trench in meters numerically does not exceed b. 这个条件意味着每次都是在蓄能开始时走向下一条线段,也就是说每条线段相当于花费了\(a+b\)的时间. bfs,用\(d_i\)表示到达线段i需要经过最少的线段条数,到达\(B\)的时候直接计算欧几里得距离. B. Vasya's Cal…
Codeforces Round #108 (Div. 2) C. Pocket Book 题意 给定\(N(N \le 100)\)个字符串,每个字符串长为\(M(M \le 100)\). 每次选择\(i, j, k\),然后交换串\(i\)和串\(j\)的长度为\(k\)的前缀. 操作可以做任意次,求最多能得到多少不同的字符串,\(modulo (10^9+7)\). 思路 相当于每个位置的可选字符为该列的不同字符的数量. 代码 C. Pocket Book D. Frames 题意 给定…
Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) 说一点东西: 昨天晚上$9:05$开始太不好了,我在学校学校$9:40$放学我呆到十点然后还要跑回家耽误时间....要不然$D$题就写完了 周末一些成绩好的同学单独在艺术楼上课然后晚上下第一节晚自习和他们在回廊里玩开灯之后再关上一片漆黑真好玩 A.Andryusha and Socks 日常煞笔提.....我竟然$WA$了一次忘了$n<<1$ #include <…
Codeforces Round #467 (div.2) 我才不会打这种比赛呢 (其实本来打算打的) 谁叫它推迟到了\(00:05\) 我爱睡觉 题解 A. Olympiad 翻译 给你若干人的成绩 让你划定一个分数线 使得所有不低于这个分数线的人都可以获奖 但是\(0\)分的人一定不能得奖 问你有多少种获奖情况 题解 \(sort+unique\) 然后判断一下最小值是不是\(0\)就行了 #include<iostream> #include<cstdio> #include…