题目链接: D1. The Wall (easy) time limit per test 0.5 seconds memory limit per test 256 megabytes input standard input output standard output "The zombies are lurking outside. Waiting. Moaning. And when they come..." "When they come?" &quo…

题意:给定一个图,问你有几个连通块. 析:不用说了,最简单的DFS. 代码如下: #include <bits/stdc++.h> using namespace std; const int maxn = 100 + 5; const int dr[] = {1, -1, 0, 0}; const int dc[] = {0, 0, 1, -1}; char a[maxn][maxn]; int vis[maxn][maxn]; void dfs(int r, int c){ vis[r][…

题意: 取一字符串不相交的前缀和后缀(可为空)构成最长回文串. 思路: 先从两边取对称的前后缀,之后再取余下字符串较长的回文前缀或后缀. #include <bits/stdc++.h> using namespace std; bool ok(const string &s,int l,int r){ while(l<=r&&s[l]==s[r]) ++l,--r; return l>r; } void solve(){ string s;cin>&…

题意:给你一个数组a,询问m次,每次返回长度为k的和最大的子序列(要求字典序最小)的pos位置上的数字. 题解:和最大的子序列很简单,排个序就行,但是题目要求字典序最小,那我们在刚开始的时候先记录每个数的位置再排序,然后选出k个最大的数后在对位置从小到大排个序就行了(这题有个坑,第一次排序的时候记得把相等的数按位置小的排在前面). 代码: 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #…

Basic wall maze Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 168    Accepted Submission(s): 52 Special Judge Problem Description In this problem you have to solve a very simple maze consisti…

题目链接:http://codeforces.com/problemset/problem/377/A 题解: 有tot个空格(输入时统计),把其中k个空格变为wall,问怎么变才能使得剩下的空格依然为连通的.把问题反过来,其实就是求tot-k的连通图.dfs:在搜索过的空格中做个标记,同时更新连通个数. 代码如下: #include<cstdio>//hdu3183 CodeForces 377A dfs #include<cstring> #include<cmath&g…

J. Deck Shuffling Time Limit: 2   Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/problem/J Description The world famous scientist Innokentiy continues his innovative experiments with decks of cards. Now he has a deck of n cards and k…

Problem A. Poetry Challenge Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100500/attachments Description Let’s check another challenge of the IBM ICPC Chill Zone, a poetry challenge. One says a poetry string that starts with a…

题目链接:http://codeforces.com/contest/580/problem/C #include<cstdio> #include<vector> #include<cstring> #define MAX 100010 using namespace std; vector <int> a[MAX]; int visit[MAX]; int cat[MAX]; int leaf[MAX]; int ans,m; void dfs(int…

题目链接:http://codeforces.com/contest/616/problem/C 题意就是 给你一个n行m列的图,让你求’*‘这个元素上下左右相连的连续的’.‘有多少(本身也算一个),每个’*‘的结果取模10.要是为’*‘输出结果,否则输出’.‘. 这个题目就是让你求连续的'.'联通块元素个数,求完一个联通块就把这个联通块标个记号(我设了ok[][]二维数组 表示这个位置的元素的联通块的标号,相连的则为同一个标号),之后这个联通块的每个元素的ans都为f(f为联通块元素个数),然…