题意:0~30000有30001个地方,每个地方有一个或多个金币,第一步走到了d,步长为d,以后走的步长可以是上次步长+1,-1或不变,走到某个地方可以收集那个地方的财富,现在问走出去(>30000)之前最多可以收集到多少财富. 解法:容易想到DP,dp[i][j]表示到达 i 处,现在步长为 j 时最多收集到的财富,转移也不难,cnt[i]表示 i 处的财富. dp[i+step-1] = max(dp[i+step-1],dp[i][j]+cnt[i+step+1]) dp[i+step]…

题目传送门 /* 题意:两点之间有不同颜色的线连通,问两点间单一颜色连通的路径有几条 DFS:暴力每个颜色,以u走到v为结束标志,累加条数 注意:无向图 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <string> #include <vector> using namespace std; ; co…

题目传送门 /* 水题:vector容器实现插入操作,暴力进行判断是否为回文串 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <string> #include <vector> using namespace std; ; const int INF = 0x3f3f3f3f; vector<c…

D. Mr. Kitayuta's Colorful Graph Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/506/problem/D Description Mr. Kitayuta has just bought an undirected graph with n vertices and m edges. The vertices of the graph are numbered…

B. Mr. Kitayuta's Colorful Graph time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mr. Kitayuta has just bought an undirected graph consisting of n vertices and m edges. The vertices of the g…

D - Mr. Kitayuta's Colorful Graph 思路:我是暴力搞过去没有将答案离线,感觉将答案的离线的方法很巧妙.. 对于一个不大于sqrt(n) 的块,我们n^2暴力枚举, 对于大于sqrt(n)的块,我们暴力枚举答案. 这样就能做到严格sqrt(n) * n #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #defin…

题目地址:http://codeforces.com/contest/506/problem/B 先用强连通判环.然后转化成无向图,找无向图连通块.若一个有n个点的块内有强连通环,那么须要n条边.即正好首尾相连形成一条环,那么有了这个环之后,在这个块内的全部要求都能实现. 假设没有强连通环,那么就是一棵树,那么仅仅须要n-1条边就可以. 代码例如以下: #include <iostream> #include <string.h> #include <math.h> #…

数据规模小,所以就暴力枚举每一种颜色的边就行了. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<algorithm>…

由于字符串的长度很短,所以就暴力枚举每一个空每一个字母,出现行的就输出.这么简单的思路我居然没想到,临场想了很多,以为有什么技巧,越想越迷...是思维方式有问题,遇到问题先分析最简单粗暴的办法,然后一步一步的优化,不能盲目的想. 这道题要AC的快需要熟悉string的各种用法.这里做个简单总结:C++中string的常见用法. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstrin…

题目链接: C. Mr. Kitayuta, the Treasure Hunter time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The Shuseki Islands are an archipelago of 30001 small islands in the Yutampo Sea. The islands are…

A. Mr. Kitayuta, the Treasure Hunter time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The Shuseki Islands are an archipelago of 30001 small islands in the Yutampo Sea. The islands are evenly…

题意:有30001个岛,在一条线上,从左到右编号一次为0到30000.某些岛屿上有些宝石.初始的时候有个人在岛屿0,他将跳到岛屿d,他跳跃的距离为d.如果当前他跳跃的距离为L,他下一次跳跃的距离只能为L-1,L,L+1之一且不能为0.他只能往编号更大的岛跳,直到他不能跳,问他最多能收集多少个宝石? 思路:用dp[i][j]表示在第i个岛,上一步跳的距离为j的收集到的最多宝石的个数.这样如果直接表示的话,j最大可能是30000,空间会超,但是所跳跃的距离不会超过d+250, 因为额1+2+3+..…

A. Mr. Kitayuta's Gift (枚举) 题意: 给一个长度不超过10的串,问能否通过插入一个字符使得新串成为回文串. 分析: 因为所给的串很多,所以可以枚举 “在哪插入” 和 “插入什么”,写一个二重循环枚举新串,判断是否为回文串.时间复杂度为O(n3) 还可只枚举插入位置(在那个位置用一个特殊字符表示),在判断的时候,如果遇到特殊字符,则所插入的字符一定为镜像的字符. #include <cstdio> #include <cstring> ], s1[]; in…

A.Mr. Kitayuta, the Treasure Hunter 很显然的一个DP,30000的数据导致使用map+set会超时.题解给了一个非常实用的做法,由于每个点有不超过250种状态,并且这些状态都是以包含d连续的一段数字,那么可以以对d的偏移量作为状态.这算是很常见的一个优化了. #include<bits/stdc++.h> using namespace std; ; ],a[INF]; , x; int main() { scanf ("%d %d",…

B. Mr. Kitayuta's Colorful Graph time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mr. Kitayuta has just bought an undirected graph consisting of n vertices and m edges. The vertices of the g…

解题思路:给出n个点,m条边(即题目中所说的两点之间相连的颜色) 询问任意两点之间由多少种不同的颜色连接 最开始想的时候可以用传递闭包或者并查集来做,可是并查集现在还不会做,就说下用传递闭包来做的这种--- 最开始想的时候用传递闭包,可是想到传递闭包只能判断两点是否连通,不能判断连通这两点的颜色是不是一样的,所以当时想再另外用一个数组来放两点之间的颜色,没有写出来---- 然后今天去翻了别人的代码,发现把传递闭包的d数组改成三维的就可以解决问题了(因为注意到n,m的值都很小,四重循环再加一个if…

Description The Shuseki Islands are an archipelago of 30001 small islands in the Yutampo Sea. The islands are evenly spaced along a line, numbered from 0 to 30000 from the west to the east. These islands are known to contain many treasures. There are…

A. Mahmoud and Longest Uncommon Subsequence time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest comm…

题目链接:CF#286 - A 这场CF就这样爆零了...我真是太蒟蒻了... 题目分析 比赛的时候看到A题就发现不会,之后一直也没想出来,于是就弃了,还好不提交也不掉Rating... 比赛后看评论,看到有人说“I could not even solve the problem A, shame on me.” ,立刻就感觉到我是多么的蒟蒻... 看了评论中有人发的题解,就一句 “Normal DP”,再看了他的简单的解释,这才恍然大悟... 这道题就可以使用普通的DP,用 f[i][j]…

题目链接 http://codeforces.com/contest/711/problem/C 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 wit…

E. Vladik and cards 题目链接 http://codeforces.com/contest/743/problem/E 题面 Vladik was bored on his way home and decided to play the following game. He took n cards and put them in a row in front of himself. Every card has a positive integer number not e…

D - Chloe and pleasant prizes 链接 http://codeforces.com/contest/743/problem/D 题面 Generous sponsors of the olympiad in which Chloe and Vladik took part allowed all the participants to choose a prize for them on their own. Christmas is coming, so sponso…

题目链接: http://www.codeforces.com/contest/675/problem/E 题意: 对于第i个站,它与i+1到a[i]的站有路相连,先在求所有站点i到站点j的最短距离之和(1<=i<j<=n) 题解: 这种所有可能都算一遍就会爆的题目,有可能是可以转化为去求每个数对最后答案的贡献,用一些组合计数的方法一般就可以很快算出来. 这里用dp[i]表示第i个站点到后面的所有站的最短距离之和,明显,i+1到a[i]的站,一步就可以到,对于后面的那些点,贪心一下,一定…

题目链接:http://codeforces.com/contest/560/problem/E 给你一个n*m的网格,有k个坏点,问你从(1,1)到(n,m)不经过坏点有多少条路径. 先把这些坏点排序一下. dp[i]表示从(1,1)到第i个坏点且不经过其他坏点的路径数目. dp[i] = Lucas(x[i], y[i]) - sum(dp[j]*Lucas(x[i]-x[j], y[i]-x[j])) , x[j] <= x[i] && y[j] <= y[i] //到i…

B. Wizards and Huge Prize Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/167/problem/B Description One must train much to do well on wizardry contests. So, there are numerous wizardry schools and magic fees. One of such ma…

Roman and Numbers time limit per test 4 seconds memory limit per test 512 megabytes input standard input output standard output Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his…

题目:http://codeforces.com/contest/1153/problem/D 题意:给你一棵树,每个节点有一个操作,0代表取子节点中最小的那个值,1代表取子节点中最大的值,叶子节点的话就是自己置一个值,有k个子节点,那么每个子节点的值范围 就是1-k,1-k只能用一次 思路:贪心不好取,我肯定是要排列完才知道当前的值是哪个,但是我可以知道当前节点应该是取子节点中排名第几的那个,从而推出根节点1的排名,然后再输出就行,因为是要从 子节点那里递归上来,所以我们采取树形dp #inc…

题目链接: http://codeforces.com/problemset/problem/258/B B. Little Elephant and Elections time limit per test2 secondsmemory limit per test256 megabytes 问题描述 There have recently been elections in the zoo. Overall there were 7 main political parties: one…

题目链接: http://codeforces.com/problemset/problem/401/D D. Roman and Numbers time limit per test4 secondsmemory limit per test512 megabytes 问题描述 Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sere…

http://codeforces.com/contest/816/problem/E 题意: 去超市买东西,共有m块钱,每件商品有优惠卷可用,前提是xi商品的优惠券被用.问最多能买多少件商品? 思路: 第一件商品使用优惠券不需要前提,别的都是需要的,然后这样就形成了一棵以1为根的树. 这样,很容易想到是树形dp. d[u][j][0/1]表示以u为根的子数中选择j件商品所需的最少花费,0/1表示u商品是否能用优惠券. 解释一下代码中的sz[],它所代表的是以u为根的子树的结点数. 当我们现在访…