CSU 1508：地图的四着色（DFS+剪枝）

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1508 题意:地图中四联通的块是一个国家,A和B每个人可以涂两种颜色,且B不能涂超过5次,相邻的国家不能涂一样的颜色,并且最后每种颜色都要用上,问共有多少种涂色方案. 思路:先DFS处理出图中的国家,然后先用一个邻接矩阵存下相邻的国家(直接用邻接表会有重边),然后再用邻接表存图.数方案个数的时候,假设共有a,b,c,d这四种颜色,A可以涂a,b,B可以涂c,d.因为颜色之间没有差异,所以A先涂a颜…

[csu1508 地图的四着色]二分图染色

抽象后的题意:给一个不超过30个点的图,A从中选不超过5个点涂红绿两种颜色,B用黑白两种颜色把剩下的涂完,任意一条边两端的颜色不同,求每种颜色至少用涂一次的方案数思路:枚举A涂的点的集合,将原图分成两个子图P和Q,P和Q互相不影响,因为涂的颜色不同.考虑A在P中涂颜色,由于一条边的两端的颜色不能相同,于是对P进行二分染色,如果是非二分图,那么方案数为0,否则令P的连通分量个数为cnt,如果点集小于2则方案总数为0,否则如果边数为0,说明给P涂1种颜色的答案为2,最后答案为2cnt-2,边数不为…

DFS+剪枝 HDOJ 5323 Solve this interesting problem

题目传送门 /* 题意:告诉一个区间[L,R],问根节点的n是多少 DFS+剪枝:父亲节点有四种情况:[l, r + len],[l, r + len - 1],[l - len, r],[l - len -1,r]; */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <queue> using namespace std;…

数据结构与算法实验题 9.1 K 歌 DFS+剪枝

数据结构与算法实验题 K 歌 ★实验任务 3* n 个人(标号1~ 3 * n )分成 n 组 K 歌.有 m 个 3 人组合,每个组合都对应一个分数,你能算出最大能够得到的总分数么? ★数据输入输入的第 1 行是一个正整数 n 和 m(1<= n <=6 , n<=m<= C(3*n,3) 组合数) ,输入的第2~m+1行每行包括四个正整数a,b,c,s(1<=a , b ,c <=3 *n, a,b,c不相等,1<=s<=10 ^8)表示组合(a,b…

*HDU1455 DFS剪枝

Sticks Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9779 Accepted Submission(s): 2907 Problem Description George took sticks of the same length and cut them randomly until all parts became…

POJ 3009 DFS+剪枝

POJ3009 DFS+剪枝原题: Curling 2.0 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16280 Accepted: 6725 Description On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from our…

poj 1724:ROADS（DFS + 剪枝）

ROADS Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10777 Accepted: 3961 Description N cities named with numbers 1 ... N are connected with one-way roads. Each road has two parameters associated with it : the road length and the toll…

DFS(剪枝) POJ 1011 Sticks

题目传送门 /* 题意:若干小木棍,是由多条相同长度的长木棍分割而成,问最小的原来长木棍的长度: DFS剪枝:剪枝搜索的好题!TLE好几次,终于剪枝完全! 剪枝主要在4和5:4 相同长度的木棍不再搜索:5 若新的搜索连第一条都没组合出来,直接break: 详细解释:http://blog.csdn.net/lyy289065406/article/details/6647960 http://www.cnblogs.com/devil-91/archive/2012/08/03/2621787.…

HDU 5952 Counting Cliques 【DFS+剪枝】（2016ACM/ICPC亚洲区沈阳站）

Counting Cliques Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 539 Accepted Submission(s): 204 Problem Description A clique is a complete graph, in which there is an edge between every pair…

HDU 5937 Equation 【DFS+剪枝】（2016年中国大学生程序设计竞赛（杭州））

Equation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 92 Accepted Submission(s): 24 Problem Description Little Ruins is a studious boy, recently he learned addition operation! He was rewa…

LA 6476 Outpost Navigation （DFS+剪枝）

题目链接 Solution DFS+剪枝对于一个走过点k,如果有必要再走一次,那么一定是走过k后在k点的最大弹药数增加了.否则一定没有必要再走. 记录经过每个点的最大弹药数,对dfs进行剪枝. #include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include <map> using namespace std; map<string…

poj 1011 Sticks （DFS+剪枝）

Sticks Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 127771 Accepted: 29926 Description George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the or…

poj 1564 Sum It Up | zoj 1711 | hdu 1548 (dfs + 剪枝 or 判重)

Sum It Up Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 4 Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Given a specified total t and…

C#的百度地图开发(四)前端显示与定位

原文:C#的百度地图开发(四)前端显示与定位有了这些定位信息,那要如何在前端的页面上显示出来呢?这需要用到百度地图的JavaScript的API.下面是示例代码. 前端代码 <%@ Page Language="C#" AutoEventWireup="true" CodeBehind="ViewMap.aspx.cs" Inherits="TEST.ViewMap" %> <!DOCTYPE html&g…

hdu 4109 dfs+剪枝优化

求最久时间即在无环有向图里求最远路径 dfs+剪枝优化从0节点(自己添加的)出发,0到1~n个节点之间的距离为1.mt[i]表示从0点到第i个节点眼下所得的最长路径 #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<vector> using namespace std; const…

PKU 2531 Network Saboteur(dfs+剪枝||随机化算法)

题目大意:原题链接给定n个节点,任意两个节点之间有权值,把这n个节点分成A,B两个集合,使得A集合中的每一节点与B集合中的每一节点两两结合(即有|A|*|B|种结合方式)权值之和最大. 标记:A集合:true B集合:false 解法一:dfs+剪枝 #include<iostream> #include<cstring> using namespace std; int n,ans; ]; ][]; void dfs(int i,int cursum) { in[i]=tru…

HDU 1501 Zipper 【DFS+剪枝】

HDU 1501 Zipper [DFS+剪枝] Problem Description Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in…

HDOJ 1501 Zipper 【DP】【DFS+剪枝】

HDOJ 1501 Zipper [DP][DFS+剪枝] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10886 Accepted Submission(s): 3925 Problem Description Given three strings, you are to determine whether the third str…

poj 1011 ：Sticks (dfs+剪枝)

题意:给出n根小棒的长度stick[i],已知这n根小棒原本由若干根长度相同的长木棒(原棒)分解而来.求出原棒的最小可能长度. 思路:dfs+剪枝.蛮经典的题目,重点在于dfs剪枝的设计.先说先具体的实现:求出总长度sum和小棒最长的长度max,则原棒可能的长度必在max~sum之间,然后从小到大枚举max~sum之间能被sum整除的长度len,用dfs求出所有的小棒能否拼凑成这个长度,如果可以,第一个len就是答案. 下面就是关键的了,就是这道题dfs的实现和剪枝的设计: 1.以一个小棒为开头…

Black And White(DFS+剪枝)

Black And White Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 3937 Accepted Submission(s): 1082Special Judge Problem Description In mathematics, the four color theorem, or the four color…

codeforces 615 B. Longtail Hedgehog (DFS + 剪枝)

题目链接: codeforces 615 B. Longtail Hedgehog (DFS + 剪枝) 题目描述: 给定n个点m条无向边的图,设一条节点递增的链末尾节点为u,链上点的个数为P,则该链的beauty值 = P*degree[u].问你所有链中最大的beauty值. 解题思路: 因为只需要找到以每个节点为终点的最长递增链即可,所以建图的时候可以建成从编号小的节点到编号大的节点的有向图,然后用DFS搜索,但是直接暴搜会TLE,然后就开始了我的TLE之路,TLE24, TLE42,TL…