每一行都是一组测试案例   第一个数字 表示总和 第二个数字表示 一共有几个可用数据  现在 按照从小到大的顺序   输出  那些数字中若干数字之和为总和的  信息 /. 很好很明显的  遍历痕迹 , 多看多练 // 利用vector不定长数组 构图 然后就知道 某个节点相邻的 所有节点 #include<stdio.h> #include<string.h> #include<math.h> #include<iostream> #include<a…

题目链接:http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=927 代码: #include <stdio.h> #include <stdlib.h> #include <string.h> #include <math.h> #include <algorithm> #include <iostream> using namespace std; int n,k; ]; ; ]…

结束了三分搜索的旅程 我开始迈入深搜的大坑.. 首先是一道比较基础的深搜题目(还是很难理解好么) POJ 1564 SUM IT UP 大体上的思路无非是通过深搜来进行穷举.匹配 为了能更好地理解深搜 可以尝试去画一下二叉树理解一下,查看遍历的路径 代码还是百度到别人的自己再参悟- -佩服别人的功底啊 先上代码: /*POJ 1546 Sum it up*/ # include<iostream> # include<algorithm> # include<cstdio&g…

链接: http://poj.org/problem?id=1564 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88230#problem/F 给了一个数 m,给一个由 n 个数组成的数组 a[] , 求和为 m 的 a[] 的子集 pos 代表的是搜到第 pos 个元素,ans 代表的是 b[] 数组中存的第 ans 个数 我一定要学会深搜!!! 代码: #include <cstdio> #include <cstri…

Description 请考虑一个由1到N(N=3, 4, 5 ... 9)的数字组成的递增数列:1 2 3 ... N. 现在请在数列中插入“+”表示加,或者“-”表示减,抑或是“ ”表示空白,来将每一对数字组合在一起(请不在第一个数字前插入符号). 计算该表达式的结果并注意你是否得到了和为零. 请你写一个程序找出所有产生和为零的长度为N的数列. Input 单独的一行表示整数N (3 <= N <= 9). Output 按照ASCII码的顺序,输出所有在每对数字间插入“+”, “-”,…

题目链接 https://acm.bnu.edu.cn/v3/problem_show.php?pid=52305 problem  description In ICPCCamp, there are n cities and (n−1) (bidirectional) roads between cities. The i-th road is between the ai-th and bi-th cities. It is guaranteed that cities are conne…

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…

POJ 2676 Sudoku Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17627   Accepted: 8538   Special Judge Description Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the…

[题目链接:HDOJ-2952] Counting Sheep Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2476    Accepted Submission(s): 1621 Problem Description A while ago I had trouble sleeping. I used to lie awake,…

http://poj.org/problem?id=3249 Test for Job Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 8206   Accepted: 1831 Description Mr.Dog was fired by his company. In order to support his family, he must find a new job as soon as possible. No…

最小生成树计数 Description 现在给出了一个简单无向加权图.你不满足于求出这个图的最小生成树,而希望知道这个图中有多少个不同的最小生成树.(如果两颗最小生成树中至少有一条边不同,则这两个最小生成树就是不同的).由于不同的最小生成树 可能很多,所以你只需要输出方案数对31011的模就可以了. Input 第 一行包含两个数,n和m,其中1<=n<=100; 1<=m<=1000; 表示该无向图的节点数和边数.每个节点用1~n的整数编号.接下来的m行,每行包含两个整数:a,…

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…

题目链接: http://codeforces.com/problemset/problem/707/D 题目大意: 一个N*M的书架,支持4种操作 1.把(x,y)变为有书. 2.把(x,y)变为没书. 3.把x行上的所有书状态改变,有变没,没变有. 4.回到第K个操作时的状态. 求每一次操作后书架上总共多少书. 题目思路: [离线][深搜][树] 现场有思路不过没敢写哈.还是太弱了. 总共只用保存一张图,把操作看成一棵树,一开始I操作连接在I-1操作后,如果遇到操作4的话,把I操作与I-1操…

题目大意:有一堆木棍 由几个相同长的木棍截出来的,求那几个相同长的木棍最短能有多短? 深搜+剪枝 具体看代码 #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <ctime> #include <algorithm> #include <iostream> #include <sstream> #i…

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…

<span style="color:#330099;">/* I - 深搜 基础 Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u Submit Status Description Given a specified total t and a list of n integers, find all distinct sums using numbers from the…

1353. Milliard Vasya's Function Time limit: 1.0 second Memory limit: 64 MB Vasya is the beginning mathematician. He decided to make an important contribution to the science and to become famous all over the world. But how can he do that if the most i…

Description Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer…

图的深搜与广搜 一.介绍: p { margin-bottom: 0.25cm; direction: ltr; line-height: 120%; text-align: justify; orphans: 0; widows: 0 } p.western { font-family: "Calibri", serif; font-size: 10pt } p.cjk { font-family: "宋体"; font-size: 10pt } p.ctl {…

Description An abandoned country has n(n≤100000) villages which are numbered from 1 to n. Since abandoned for a long time, the roads need to be re-built. There are m(m≤1000000) roads to be re-built, the length of each road is wi(wi≤1000000). Guarante…

传送门:http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=4976 时间限制(普通/Java):1000MS/3000MS     内存限制:65536KByte 描述 我们定义十进制数中有6的为新生数.现在给你一些二进制数,叫你判断是不是新生数.当然这个太简单了,我们换种玩法,给你一个不完全的二进制数,其中有些数字是未知的用x表示,问,这个未知的二进制数最多可以有几种新生数的表示. 输入 输入有多组数…

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>…

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…

 Stamps  The government of Nova Mareterrania requires that various legal documents have stamps attached to them so that the government can derive revenue from them. In terms of recent legislation, each class of document is limited in the number of st…

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…

一门武功能否传承久远并被发扬光大,是要看缘分的.一般来说,师傅传授给徒弟的武功总要打个折扣,于是越往后传,弟子们的功夫就越弱-- 直到某一支的某一代突然出现一个天分特别高的弟子(或者是吃到了灵丹.挖到了特别的秘笈),会将功夫的威力一下子放大N倍 -- 我们称这种弟子为"得道者". 这里我们来考察某一位祖师爷门下的徒子徒孙家谱:假设家谱中的每个人只有1位师傅(除了祖师爷没有师傅):每位师傅可以带很多徒弟:并且假设辈分严格有序,即祖师爷这门武功的每个第i代传人只能在第i-1代传人中拜1个师…

http://acm.hdu.edu.cn/showproblem.php?pid=2510 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 729    Accepted Submission(s): 361 Problem Description 符号三角形的 第1行有n个由“+”和”-“组成的符号 ,以后每行符号比上行少1个,2个…

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18201   Accepted: 9192 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100…

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…

个人心得:对于深搜的使用还是不到位,对于递归的含义还是不太清楚!本来想着用深搜构成一个排列,然后从一到n分割成俩个数组,然后后面发现根本实现不了,思路太混乱.后来借鉴了网上的思想,发现用数组来标志,当值等于一时就属于A数组,等于0时属于B数组,这样就可以构成递归,即下一个数只有在A数组和不在A数组,发现这个思想真的挺好的. DFS心得:若从第二步开始,动作变得程序化,像地图,只有上下左右,这个是只有为0不为0是就可以用深搜递归思想解决. A university network is compo…