题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5319 Painter Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 826    Accepted Submission(s): 383 Problem Description Mr. Hdu is an painter, as we al…

Problem Description Given a rectangle frame of size n×m. Initially, the frame is strewn with n×m square blocks of size 1×1. Due to the friction with the frame and each other, the blocks are stable and will not drop. However, the blocks can be knocked…

Problem Description === Op tech briefing, 2002/11/02 06:42 CST === "The item is locked in a Klein safe behind a painting in the second-floor library. Klein safes are extremely rare; most of them, along with Klein and his factory, were destroyed in Wo…

Square Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9588    Accepted Submission(s): 3127 Problem Description Given a set of sticks of various lengths, is it possible to join them end-to-end…

Painter 题意:有一个棋盘n行,列数不超过50,用red和blue给这个棋盘涂色,每个格子每种颜色最多涂一次,如果两种颜色都涂了则该格子颜色为Green;red以斜杠'\'方式涂色,bule以'/'方式涂色.给出涂色后的棋盘,问最少涂了几次. 思路:搜索的思路没错,不过这题有点小坑,题目只告诉了行,并没有明确列,所以得自己求出.涂色时并不是一涂到底,也可以连续几个对角的格子涂色.所以不管是暴力枚举还是搜索都行. const int N=100+5; char s[N][N]; int n,…

Painter Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 745    Accepted Submission(s): 345 Problem Description Mr. Hdu is an painter, as we all know, painters need ideas to innovate , one day,…

题意: 一个画家画出一张,有3种颜色的笔,R.G.B.R看成'\',B看成'/',G看成这两种的重叠(即叉形).给的是一个矩阵,矩阵中只有4种符号,除了3种颜色还有'.',代表没有涂色.问最小耗费多少笔即可画成这副图? 思路: 最小耗费就是斜着的可以一笔搞定,但是如果中间隔着'.'或者其他一种形状,则不能一笔,要变两笔.主要麻烦在矩阵不是正方形,而可能是长方形.其实只要按照其画法,从左上往右下方向画,逐个条斜线扫描即可.另一个方向'/'也是如此. 我看到模拟本来就不想敲,扫两遍矩阵,用了vect…

题意:红色从左上向右下涂,蓝色从右上向左下涂,既涂红色又涂蓝色就变成绿色,问最少涂几下能变成给的图. 解法:模拟一下就好了,注意细节. 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h> #include<math.h> #include<limits.h> #include<…

这道题目还是比较水的,但是题意理解确实费了半天劲,没办法 谁让自己是英渣呢! 题目大意: 猪脚要解决问题, 他有个习惯,每次只解决比之前解决过的问题的难度要大. 他给我们一个矩阵  矩阵的 i 行 j 列表示 解决完第 i 个问题后再解决第 j 个问题 花费时间为 T[i][j] 也就是 题目的难度. 并且他是从第0个问题开始解决的,第0个问题花费的时间为 0 下面是代码 : #include<stdio.h> #include<stdlib.h> #include<math…

题目链接 题意 求\(\sum_{d|n}\phi (d) \times {n\over d}\),其中\(\phi(n) = n\prod_{p|n}({1-{1\over p}})\) 分析 将\(\phi(d)\) 分解式子代入可知:\(\sum_{d|n}(n\times \prod_{p|d}(1-{1\over p}))\) \(d\) 是 \(n\) 的因子,枚举 \(d\) 的质因子的所有可能的组成情况共\(2^c\)中. 其中 c 为 n 的不同质因子个数(即题目中输入的 n…

HDU 1241 是深搜算法的入门题目,递归实现. 原题目传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1241 代码仅供参考,c++实现: #include <iostream> using namespace std; ][]; int p,q; void dfs(int x,int y){ land[x][y] = '*'; ][y]!= ][y] != ] != ] != ][y+]!= ][y-] != ][y-] != ][y+] !…

Rikka with Tree 题意:给出树的定义,给出树相似的定义和不同的定义,然后给出一棵树,求是否存在一颗树即和其相似又与其不同.存在输出NO,不存在输出YES. 思路:以1号节点为根节点,我们观察到一颗树如果不存在这种树即与其相似又与其不同,那么这棵树要么所有节点的深度都不一样,要么有深度一样的并且他们的父节点都一样,后来想想,如果当前节点的父节点有多个儿子,那么当前节点则不能有儿子节点,否则就可以将儿子节点转移到兄弟节点上.思路很正确,比赛的时候用深搜递归写就错了,改成两层循环枚举任意…

#include<stdio.h>#include<string.h>char mapp[220][220];int m,n,mmin;void dfs(int x,int y,int time){  if(mapp[x][y]=='#'){   return ;  }  if(x<0||x>=m||y<0||y>=n){   return ;  }  if(mapp[x][y]=='a'){         dfs(x,y+1,0);         df…

题意: 给出老虎的起始点.方向和驴的起始点.方向.. 规定老虎和驴都不会走自己走过的方格,并且当没路走的时候,驴会右转,老虎会左转.. 当转了一次还没路走就会停下来.. 问他们有没有可能在某一格相遇.. 思路: 模拟,深搜.. 用类似时间戳的东西给方格标记上,表示某一秒正好走到该方格.. 最后遍历一下驴在某一格方格标记时间是否和老虎在该格标记的时间一样,一样代表正好做过这里了.. 还有一种情况就是老虎或驴一直停在那里,那就算不相等,也是可以的.. Tips: 我一直忘了老虎或驴停下来的情况,这样…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5648 题意:给定n,m(1<= n,m <= 15,000),求Σgcd(i|j,i&j);(1 <= i <= n,1<=j<=m); 至多三组数据,至多两组数据max(n,m) > 2000.至多一组数据max(n,m) > 8000; 很多题解是用递推打表,将数据压缩250倍,即[i][j]:代表[1...250*i][1...250*j],之后零…

N皇后问题 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1757    Accepted Submission(s): 772   Problem Description 在N*N的方格棋盘放置了N个皇后,使得它们不相互攻击(即任意2个皇后不允许处在同一排,同一列,也不允许处在与棋盘边框成45角的斜线上.你的任务是,对于给定的N,求出…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=1241 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18758    Accepted Submission(s): 10806 Problem Description The GeoSurvComp geologic survey comp…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1518 题目大意:根据题目所给的几条边,来判断是否能构成正方形,一个很好的深搜应用,注意剪枝,以防超时! #include <iostream> #include <cstdio> #include<algorithm> #include <cstring> using namespace std; ],visit[]; int l,n; int dfs(int…

题目 //传说中的记忆化搜索,好吧,就是用深搜//多做题吧,,这个解法是搜来的,蛮好理解的 //题目大意:给出两堆牌,只能从最上和最下取,然后两个人轮流取,都按照自己最优的策略,//问说第一个人对多的分值.//解题思路:记忆化搜索,状态出来就非常水,dp[fl][fr][sl][sr][flag],//表示第一堆牌上边取到fl,下面取到fr,同样sl,sr为第二堆牌,flag为第几个人在取.//如果是第一个人,dp既要尽量大,如果是第二个人,那么肯定尽量小. http://www.2cto.co…

Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task is: approa…

DES:从23个队员中选出4—4—2—1共4种11人来组成比赛队伍.给出每个人对每个职位的能力值.给出m组人在一起时会产生的附加效果.问你整场比赛人员的能力和最高是多少. 用深搜暴力枚举每种类型的人选择情况.感觉是这个深搜写的很机智. 在vector中存结构体就会很慢.TLE.直接存序号就AC了.以后还是尽量少用结构体吧. #include<stdio.h> #include<string.h> #include<map> #include<vector>…

THE MATRIX PROBLEM Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5437    Accepted Submission(s): 1372 Problem Description You have been given a matrix CN*M, each element E of CN*M is positive…

Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10541    Accepted Submission(s): 3205Special Judge Problem Description The Princess has been abducted by the BEelzebub…

题目链接 Problem Description 在N*N的方格棋盘放置了N个皇后,使得它们不相互攻击(即任意2个皇后不允许处在同一排,同一列,也不允许处在与棋盘边框成45角的斜线上. 你的任务是,对于给定的N,求出有多少种合法的放置方法. Input 共有若干行,每行一个正整数N≤10,表示棋盘和皇后的数量:如果N=0,表示结束.   Output 共有若干行,每行一个正整数,表示对应输入行的皇后的不同放置数量.   Sample Input 1 8 5 0   Sample Output 1…

中文题 深搜 许久没写鸟,卡在输入问题上... #include <iostream> #include <string> using namespace std; bool flg; int n; ][]; ]; void dfs(int p) { int i; if (p >=n || flg) return; ] == 'm') { flg = true; return; } ; i<n; i++) ] == str[p][]) { used[i] = true;…

题目 这是传说中的深搜吗....不确定,,,,貌似更加像是模拟,,,, //我要做深搜题目拉 //实际上还是模拟 #include<iostream> #include<string> #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> using namespace std; int ans; ]; __int64 n; ]; int l…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 149833    Accepted Submission(s): 39945 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

题目链接:HDU 5355 http://acm.split.hdu.edu.cn/showproblem.php?pid=5355 Problem Description There are m soda and today is their birthday. The 1-st soda has prepared n cakes with size 1,2,…,n. Now 1-st soda wants to divide the cakes into m parts so that th…

深搜,从一点向各处搜找到全部能走的地方. Problem Description There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on…

Problem D: Servicing stations A company offers personal computers for sale in N towns (3 <= N <= 35). The towns are denoted by 1, 2, ..., N. There are direct routes connecting M pairs from among these towns. The company decides to build servicing st…