D. Igor In the Museum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Igor is in the museum and he wants to see as many pictures as possible. Museum can be represented as a rectangular fiel…

Igor In the Museum Descriptions 给你一个n*m的方格图表示一个博物馆的分布图.每个方格上用'*'表示墙,用'.'表示空位.每一个空格和相邻的墙之间都有一幅画.(相邻指的是上下左右相邻).你可以从一个空格的位置走到相邻的空格位置.现在你给你若干个(xi,yi)形式的询问,表示你现在在(xi,yi)这个位置(保证为空位)出发,问你从这个点出发你能看到多少幅画. Input 第一行有3个整数n,m,k(3<=n,m<=1000,1<=k<=min(m*m,…

[链接] 我是链接,点我呀:) [题意] 题意 [题解] 同一个联通块里面答案都一样. 把每个联通块的答案都算出来 然后赋值就好 [代码] #include <bits/stdc++.h> using namespace std; const int N = 1000; int n,m,k; int dx[4] = {0,0,1,-1}; int dy[4] = {1,-1,0,0}; char s[N+10][N+10]; bool bo[N+10][N+10]; int ans[N+10]…

D. Igor In the Museum Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/598/problem/D Description Igor is in the museum and he wants to see as many pictures as possible. Museum can be represented as a rectangular field of n ×…

 Igor In the Museum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Igor is in the museum and he wants to see as many pictures as possible. Museum can be represented as a rectangular field…

D. Igor In the Museum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Igor is in the museum and he wants to see as many pictures as possible. Museum can be represented as a rectangular fiel…

题目链接:http://codeforces.com/problemset/problem/598/D 题目分类:dfs 题目分析:处理的时候一次处理一片而不是一个,不然会超时 代码: #include<bits/stdc++.h> using namespace std; int n,m,k,a,b,ii; int ans; ][]; ]={,,,-}; ]={,-,,}; ][]; ]; void dfs(int x,int y) { ||y>m||y<) return ; i…

http://codeforces.com/problemset/problem/598/D 分析:BFS,同一连通区域的周长一样,但查询过多会导致TLE,所以要将连通区域的答案储存,下次查询到该连通区域就可以直接得出结果 1 #include<iostream> 2 #include<sstream> 3 #include<cstdio> 4 #include<cstdlib> 5 #include<string> 6 #include<…

题目链接:http://codeforces.com/problemset/problem/598/D 题意是 给你一张行为n宽为m的图 k个询问点 ,求每个寻问点所在的封闭的一个上下左右连接的块所能看到的壁画有多少(大概这样吧). 我的做法是bfs(dfs也可以)这个为'.'的点,要是遇到上下左右其中有'*'的话就加起来.要是每次询问然后bfs一下肯定超时,所以我用一个ans[1005][1005]记录每个点的值,ok[1005][1005]数组判断是否访问过,初始化为false.然后开始遍历…

http://codeforces.com/problemset/problem/731/C 题意:有n只袜子,m天,k个颜色,每个袜子有一个颜色,再给出m天,每天有两只袜子,每只袜子可能不同颜色,问要让每天的袜子是相同颜色的,要重新染色的袜子数最少是多少. 思路:并查集合并,将同一天的袜子合并起来,然后就形成了cnt个集合,每个集合都是独立的,因此排序,找出每个集合里面袜子颜色相同的最多的是哪个颜色,然后把其他不属于这个颜色的都染成这个颜色,那么这样重新染色的袜子数是最少的.然后每个集合的答案…