On some special occasions Nadia’s company provide very special lunch for all employees of the company. Before the food is served all of the employees must stand in a queue in front of the food counter. The company applied a rule for standing in the q…

UVA.548 Tree(二叉树 DFS) 题意分析 给出一棵树的中序遍历和后序遍历,从所有叶子节点中找到一个使得其到根节点的权值最小.若有多个,输出叶子节点本身权值小的那个节点. 先递归建树,然后DFS求解. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <string> #include <sstre…

UVA - 11853 思路:dfs,从最上面超过上边界的圆开始搜索,看能不能搜到最下面超过下边界的圆. 代码: #include<bits/stdc++.h> using namespace std; ; double l,r; int n; bool vis[N]={false}; bool flag=false; struct point { int x,y,r; }a[N]; bool intersect(point a,point b) { return (a.x-b.x)*(a.x…

UVA.540 Team Queue (队列) 题意分析 有t个团队正在排队,每次来一个新人的时候,他可以插入到他最后一个队友的身后,如果没有他的队友,那么他只能插入到队伍的最后.题目中包含以下操作: 1.ENQUEUE x :表示编号为x的入队: 2.DEQUEUE:长队的队首出队. 3.STOP:停止模拟 并且对于每一个DEQUEUE操作,输出队首的编号. 如果我们直接用一个队列来模拟的话,是无法实现的,原因在于,我们无法向队列中间插入元素.那么题目中还有一条重要的性质,那么就是:可以插入到…

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=513 终于开始接触图了,恩,开始接触DFS了,这道题就是求连通分量,比较简单. #include<iostream> #include<cstring> using namespace std; int m, n; //记录连通块的数量 ][]; ][]; void…

题目链接:uva 12253 - Simple Encryption 题目大意:给定K1.求一个12位的K2,使得KK21=K2%1012 解题思路:按位枚举,不且借用用高速幂取模推断结果. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; const ll ite=(1<<20)-1; ll N; /* l…

题目大意:给出一个方格矩阵,矩阵中有数字0~9,任选一个格子为起点,将走过的数字连起来构成一个数,找出最大的那个数,每个格子只能走一次. 题目分析:DFS.剪枝方案:在当前的处境下,找出所有还能到达的点的个数,若当前数字的长度加上个数仍小于目前最优答案的长度,则剪去:若长度相等,则将所有还能到达的数字按从大到小排序后连到当前数字上,如果还比目前最优解小,则减去.找出所有还能到达的点的过程用BFS实现. #pragma comment(linker, "/STACK:1024000000,1024…

Description In the ``Four Color Map Theorem" was proven with the assistance of a computer. This theorem states that every map can be colored using only four colors, in such a way that no region is colored using the same color as a neighbor region. He…

UVA - 1103Ancient Messages In order to understand early civilizations, archaeologists often study texts written in ancient languages.One such language, used in Egypt more than 3000 years ago, is based on characters called hieroglyphs.Figure C.1 shows…

题意:给你中序后序 求某叶子节点使得从根到该节点权值和最小.若存在多个,输出其权值最小的那个. 题解:先建树,然后暴力dfs/bfs所有路径,取min 技巧:递归传参数,l1,r1,l2,r2, sum,root, 代码: #define _CRT_SECURE_NO_WARNINGS #include "stdio.h" #include<stdio.h> #include<algorithm> #include<string> #include&…