题意:问有多少种组合方法让每一行每一列最小值都是1 思路:我们可以以行为转移的状态 附加一维限制还有多少列最小值大于1 这样我们就可以不重不漏的按照状态转移 但是复杂度确实不大行(减了两个常数卡过去的...) #include <bits/stdc++.h> using namespace std; const int inf = 0x3f3f3f3f; const double eps = 1e-6; const int N = 3e5+7; typedef long long ll; co…

链接: https://codeforces.com/contest/1228/problem/B 题意: Suppose there is a h×w grid consisting of empty or full cells. Let's make some definitions: ri is the number of consecutive full cells connected to the left side in the i-th row (1≤i≤h). In partic…

Codeforces Round #589 (Div. 2)-E. Another Filling the Grid-容斥定理 [Problem Description] 在\(n\times n\)的格子中填入\([1,k]\)之间的数字,并且保证每一行至少有一个\(1\),每一列至少有一个\(1\),问有多少种满足条件的填充方案. [Solution] 令\(R[i]\)表示为第\(i\)行至少有一个\(1\)的方案数,\(C[i]\)表示第\(i\)列至少有一个\(1\)的方案数.则题目要…

Hard problem 题目链接: http://codeforces.com/contest/706/problem/C Description Vasiliy is fond of solving different tasks. Today he found one he wasn't able to solve himself, so he asks you to help. Vasiliy is given n strings consisting of lowercase Engl…

目录 Contest Info Solutions A. Distinct Digits B. Filling the Grid C. Primes and Multiplication D. Complete Tripartite E. Another Filling the Grid Contest Info Practice Link Solved A B C D E F 5/6 O O O O Ø - O 在比赛中通过 Ø 赛后通过 ! 尝试了但是失败了 - 没有尝试 Solutions…

https://codeforces.com/contest/1228/problem/A A. Distinct Digits 超级简单嘻嘻,给你一个l和r然后寻找一个数,这个数要满足的条件是它的每一位的数字不相同,找出满足要求的最小的那个数输出,没有找到就输出-1: #include<bits/stdc++.h> using namespace std; bool check(int n){ ]={}; while(n){ ]) vis[n%]=; else return false; n…

链接: https://codeforces.com/contest/1228/problem/E 题意: You have n×n square grid and an integer k. Put an integer in each cell while satisfying the conditions below. All numbers in the grid should be between 1 and k inclusive. Minimum number of the i-t…

链接: https://codeforces.com/contest/1182/problem/A 题意: You have a given integer n. Find the number of ways to fill all 3×n tiles with the shape described in the picture below. Upon filling, no empty spaces are allowed. Shapes cannot overlap. This pict…

Is that a kind of fetishism? No, he is objectively a god. 见识了一把 Mcdic 究竟出题有多神. (虽然感觉还是吹过头了) 开了场 Virtual 玩. 开场先秒了 AB.C 居然差点没做出来,有点耻辱. 开 D.怎么不会--Div. 2 的 D 都能卡住我,我心态崩了. 调到 E. woc 这不是 sb 题吗-- 回来肝 D.想了想口胡出来了,然而心态已经崩了,用了很长很长时间才打出来. 最后只剩 20min 时开 F.这--辣鸡三合…

链接: https://codeforces.com/contest/1228/problem/D 题意: You have a simple undirected graph consisting of n vertices and m edges. The graph doesn't contain self-loops, there is at most one edge between a pair of vertices. The given graph can be disconne…

链接: https://codeforces.com/contest/1228/problem/C 题意: Let's introduce some definitions that will be needed later. Let prime(x) be the set of prime divisors of x. For example, prime(140)={2,5,7}, prime(169)={13}. Let g(x,p) be the maximum possible int…

链接: https://codeforces.com/contest/1228/problem/A 题意: You have two integers l and r. Find an integer x which satisfies the conditions below: l≤x≤r. All digits of x are different. If there are multiple answers, print any of them. 思路: 水题. 代码: #include…

[链接] 我是链接,点我呀:) [题意] 题意 [题解] 其实这道题感觉有点狗. 思路大概是这样 先让所有的点都在1集合中. 然后随便选一个点x,访问它的出度y 显然tag[y]=2 因为和他相连了嘛 然后其他没有和x相连的点显然只能和x在同一个集合中 所以其他1集合的点你会发现你想改也没法改,就算他们有可能连在一起也没用,因为你不可能再把其他的1改成2了,因为会和你之前选的A冲突(和这个你想 改的2没有边相连) 这就是这题的主要ideal,就是抓住这一点做文章. 然后接着,我们仍然是随便找一个…

#define HAVE_STRUCT_TIMESPEC#include<bits/stdc++.h>using namespace std;int f[257],fac[257],ifac[257];const long long mod = 1e9+7;int qpow(int x,int y){ int tamp=1; while(y){ if(y&1) tamp=1ll*tamp*x%mod; x=1ll*x*x%mod; y>>=1; } return tamp;…

#define HAVE_STRUCT_TIMESPEC#include<bits/stdc++.h>using namespace std;vector<int>adj[100007];map<vector<int>,int>mp;int main(){ int n,m; scanf("%d%d",&n,&m); int u,v; for(int i=1;i<=m;++i){ scanf("%d%d&qu…

题意:给你n个点 和 m条边 问是否可以分成三个集合 使得任意两个集合之间的任意两个点都有边 思路:对于其中一个集合v1 我们考虑其中的点1 假设点u和1无边 那么我们可以得到 u一定和点1在一个集合 否则将输出-1 所以我们可以暴力把所有的点先尽可能的分类 最后判断是否可行 #include <bits/stdc++.h> using namespace std; const int inf = 0x3f3f3f3f; const double eps = 1e-6; const int N…

Coloring Trees Problem Description: ZS the Coder and Chris the Baboon has arrived at Udayland! They walked in the park where n trees grow. They decided to be naughty and color the trees in the park. The trees are numbered with integers from 1 to n fr…

题目链接: http://codeforces.com/problemset/problem/268/E E. Playlist time limit per test 1 secondmemory limit per test 256 megabytes 问题描述 Manao's friends often send him new songs. He never listens to them right away. Instead, he compiles them into a play…

题目链接: http://codeforces.com/contest/464/problem/C J. Substitutes in Number time limit per test 1 secondmemory limit per test 256 megabytes 问题描述 Andrew and Eugene are playing a game. Initially, Andrew has string s, consisting of digits. Eugene sends A…

D. Fox And Jumping 题目连接: http://codeforces.com/contest/510/problem/D Description Fox Ciel is playing a game. In this game there is an infinite long tape with cells indexed by integers (positive, negative and zero). At the beginning she is standing at…

B. Maximum Submatrix 2 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/375/B Description You are given a matrix consisting of digits zero and one, its size is n × m. You are allowed to rearrange its rows. What is…

题目链接: http://codeforces.com/contest/677/problem/D 题意: 让你求最短的从start->...->1->...->2->...->3->...->...->p的最短路径. 题解: 这题dp的阶段性还是很明显的,相同的值得方格为同一个阶段,然后求从阶段1->2->3...->p的阶段图最短路. 初始化所有a[x][y]==1的格子为起始点到(x,y)坐标的距离. 方程式为dp[x1][y1…

题目链接: http://codeforces.com/contest/378/problem/E 题意: dota选英雄,现在有n个英雄,m个回合,两支队伍: 每一回合两个选择: b 1,队伍一ban掉一个英雄,或是跳过这个回合 p 1,队伍1选一个英雄 现在每个队伍都是最优决策,问最后队伍一的英雄实力和-队伍二的英雄实力和. 题解: 状压dp,决策要倒过来做,也就是dp出后面回合的决策的基础上设计出对自己当前回合对自己最有利的决策.. 注意:只有top m的英雄有可能被选,其他的都不可能被选…

题目链接:http://codeforces.com/problemset/problem/429/B 给你一个矩阵,一个人从(1, 1) ->(n, m),只能向下或者向右: 一个人从(n, 1) ->(1, m),只能向上或者向右.必须有一个相遇点, 相遇点的值不能被取到, 问两个人能得到的最大路径和是多少? dp[i][j]:表示从一个点出发的最大值:先预处理从(1,1) (1,m) (n,1) (n,m)四个点出发的4个dp最大值.然后枚举所有的点,但是这个点不能在边缘,考虑枚举点不够…

B. Strip Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/487/problem/B Description Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right. Now Alexandra wants to split it into some pieces (p…

E. LIS of Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/486/problem/E Description The next "Data Structures and Algorithms" lesson will be about Longest Increasing Subsequence (LIS for short) of a sequence.…

D. Duff in Beach Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/588/problem/D Description While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. Th…

B. Modulo Sum Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/577/problem/B Description You are given a sequence of numbers a1, a2, ..., an, and a number m. Check if it is possible to choose a non-empty subsequence aij such…

http://codeforces.com/contest/459/problem/E 不明确的是我的代码为啥AC不了,我的是记录we[i]以i为结尾的点的最大权值得边,然后wa在第35  36组数据 然后參考答案了,然后----网上一份题解 大意: 给出一个带权有向图,求经过的边权绝对上升的最长路径(可能是非简单路径,就可以能经过一个点多次)所包括的边数. 题解: 对边按权值排序后,从小到大搞. 设q[x]为已经搞过的边组成的以x点为终点的最长路径包括的边数. 设当前边e[i]为从u到v的边,…

题目:http://codeforces.com/problemset/problem/429/B 第一个人初始位置在(1,1),他必须走到(n,m)只能往下或者往右 第二个人初始位置在(n,1),他必须走到(1,m)只能往上或者往右 每个点都有个权值,要求两个人中间相遇一次且只有一次,相遇的那个点的权值不算,两个人的速度可以不一样,然后求出最大值是多少 思路:他的要求是相遇一次且只有一次,那么画几个图其实就只有两个情况了(这图是借了一位大佬的,[小声bb]) 既然知道了这个图分为了这四个区域,…