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…

链接: http://poj.org/problem?id=1321 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28899   Accepted: 14307 Description 在一个给定形状的棋盘(形状可能是不规则的)上面摆放棋子,棋子没有区别.要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列,请编程求解对于给定形状和大小的棋盘,摆放k个棋子的所有可行的摆放方案C. Input 输入含有多组测试数据.…

Problem Description Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t=4, n=6, and the list is [4,3,2,2,1,1], then there are four different sums that equal 4: 4,3…

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…

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…

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…

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…

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“+”, “-”,…

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.zju.edu.cn/onlinejudge/showProblem.do?problemId=85 http://poj.org/problem?id=1130 这两个题只有输入输出形式不一样.ZOJ的麻烦一点,这里用的ZOJ的输入输出方式 解题报告: 1.输入方式sscanf(line,"%d%d",&a,&b);表示在文本line中提取两个整形数据到a,b中 2.从外星人的角度来看,就是搜索单源最短路径,采用的方式是广搜. 3.删…

题目链接:http://poj.org/problem?id=2488 思路:按照一定的字典序深搜,当时我的想法是把所有的可行的路径都找出来,然后字典序排序. 后来,凡哥说可以在搜索路径的时候就按照字典序搜索,这样一找到可行的路径就输出来就行了.这里我吸取了之前八皇后问题时犯的错,并且优化了一下写法,就是flag,这是参考了jhf大神的写法了. 但是jhf大神的写法,思路和我一样,但是他的x,y坐标还要转来转去,我就没有这么写了,还是按照我的代码风格好一些. #include <stdio.h>…

题意: 第一行输入N M C ,表示从1到N有M条无向边,现在要从1走到N 走C次完全不同的路径,求最长边的最小值.下面M行是从a点到b点的距离. 建图: 题上说从两点之间可以有多条边,问的是从1~N的C种走法,所走路径上的最大边最小可以是多少,所以我们用结构体来储存点的距离,用二分搜索中的mid来假设成最大边的值,那么其他边都要小于等于它,然后根据mid建图,两点之间的边数作为容量网络的边权,也就是两点间的容量,(Dinic跑一遍)求出一个最大流,看它与C的关系,如果C大于它,说明mid太小,…

/* (⊙v⊙)嗯 貌似是一个建图 拓扑+深搜的过程.至于为什么要深搜嘛..一个月前敲得题现在全部推了重敲,于是明白了.因为题意要求如果有多个可能的解的话. * 就要输出字典序最小的那个.所以可以对26个英文字母从小到大尝试能否排出结果.于是出现了 深搜回溯.先选定入度为0的边框.标记为已用.将所有与它连通的 * 边框入度减一.然后递归搜索下一个.此时开始回溯.当前边框标记为未用,所有与它连通的边框入度加1. * 建图的过程则是.对每一个字母的边框测量出来.然后对A里的其它字母B.(B 覆盖 A…

迷宫问题 Description 定义一个二维数组: int maze[5][5] = { 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, }; 它表示一个迷宫,其中的1表示墙壁,0表示可以走的路,只能横着走或竖着走,不能斜着走,要求编程序找出从左上角到右下角的最短路线. Input 一个5 × 5的二维数组,表示一个迷宫.数据保证有唯一解. Output 左上角到右下角的最短路径,格式如样例所…

Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 59468   Accepted: 24750 Description Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the…

A Knight's Journey Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 35342   Accepted: 12051 Description Background  The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey …

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…

1.POJ 1564 Sum It Up 2.总结: 题意:在n个数里输出所有相加为t的情况. #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #define F(i,a,b) for (int i=a;i<=b;i++) #define mes(a,b) memse…

[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,…

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…

Curling 2.0 Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 30   Accepted Submission(s) : 16 Problem Description On Planet MM-21, after their Olympic games this year, curling is getting popular.…

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…

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…