题意: 给你两个空瓶子,只有三种操作 一.把一个瓶子灌满 二.把一个瓶子清空 三.把一个瓶子里面的水灌到另一个瓶子里面去(倒满之后要是还存在水那就依然在那个瓶子里面,或者被灌的瓶子有可能没满) 思路:BFS,打印路径时需技巧. //G++ 840K 0MS #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <stack> #in…

A. Shortest path of the king time limit per test 1 second memory limit per test 64 megabytes input standard input output standard output The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has busi…

题目是给你起点sx,和终点gx:牛在起点可以进行下面两个操作: 步行:John花一分钟由任意点X移动到点X-1或点X+1. 瞬移:John花一分钟由任意点X移动到点2*X. 你要输出最短步数及打印路径. 最短步数用bfs就行了. 至于路径,用一个结构体就可以去存每个点的父节点,再递归输出路径就行了. #include<stdio.h>#include<algorithm>#include<string.h>#include<queue>using names…

本题大意:给定一个迷宫,让你判断是否能从给定的起点到达给定的终点,这里起点需要输入起始方向,迷宫的每个顶点也都有行走限制,每个顶点都有特殊的转向约束...具体看题目便知... 本题思路:保存起点和终点的状态,保存每个顶点的状态,包括每个方向的可转向方向,然后直接BFS即可,记得保存每个儿子结点的爹,便于回溯输出路径. 参考代码: #include <bits/stdc++.h> using namespace std; struct Node{ int r, c, dir;// 表示结点的横纵…

Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18847    Accepted Submission(s): 6090Special Judge Problem Description The Princess has been abducted by the BEelzebub…

Pots Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9963   Accepted: 4179   Special Judge Description You are given two pots, having the volume of A and B liters respectively. The following operations can be performed: FILL(i)        fi…

链接: https://vjudge.net/problem/UVA-10480 题意: The regime of a small but wealthy dictatorship has been abruptly overthrown by an unexpected rebellion. Because of the enormous disturbances this is causing in world economy, an imperialist military super…

Problem Description The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrint…

问题描述 解决方法 1.像第一个问题那就是最短路问题(我代码采用迪杰斯特拉算法)实现 2.换乘次数最少,那就用bfs广搜来寻找答案.但是我的代码不能保证这个最少换乘是最短路程 代码 1 #include<stdio.h> 2 #include<iostream> 3 #include<algorithm> 4 #include<string.h> 5 #include<queue> 6 #include<vector> 7 using…

题目传送门 题意:求单词的最长公共子序列,并要求打印路径 分析:LCS 将单词看成一个点,dp[i][j] = dp[i-1][j-1] + 1 (s1[i] == s2[j]), dp[i][j] = max (dp[i-1][j], dp[i][j-1]) 代码: #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <strin…