一句话题意:在8 * 8的棋盘上,输出用最少步数从起点走到终点的方案 数据很小,可以广搜无脑解决 定义数据结构体 struct pos{ int x,y,s; //x.y表示横纵坐标,s表示步数 ]; //存储每一步的方案 }; 移动时新旧状态传递 pos u=q.front(); q.pop(); ;i<;i++) { pos th; th.x=u.x+dx[i]; th.y=u.y+dy[i]; th.s=u.s+; ;i<=u.s;i++) th.move[i]=u.move[i]; t…

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…

A. Shortest path of the king 题目连接: http://www.codeforces.com/contest/3/problem/A Description The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he…

The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, h…

题目让我们维护一个连通无向图,边有边权,支持加边删边和询问从\(x\)到\(y\)的异或最短路. 考虑到有删边这样的撤销操作,那么用线段树分治来实现,用线段树来维护询问的时间轴. 将每一条边的出现时间段标记到线段树上,表示在这一段询问中这条边存在. 异或最短路的处理方法与最大XOR和路径类似,给线段树上每个节点开一个线性基,找出当前所有的环,插入该节点的线性基,查询时任意找出一条从\(x\)到\(y\)的路径,到线性基中查询,即为答案. 具体实现时用可撤销并查集,采用按秩合并来优化,因为路径压缩…

必须要抄袭一下这个代码 The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting…

The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, h…

发一下牢骚和主题无关: 搜索,最短路都可以     每日一道理 人生是洁白的画纸,我们每个人就是手握各色笔的画师:人生也是一条看不到尽头的长路,我们每个人则是人生道路的远足者:人生还像是一块神奇的土地,我们每个人则是手握农具的耕耘者:但人生更像一本难懂的书,我们每个人则是孜孜不倦的读书郎. #include<stdio.h> #include<string.h> #include<queue> #define inf 0x3fffffff using namespace…

给你一个的棋盘, 问:从一个坐标到达另一个坐标需要多少步? 每次移动可以是八个方向.   #include <iostream> #include <cmath> #include <algorithm> #include <string> #include <cstring> #include <cstdio> #include <vector> #include <cstdlib> using namesp…

标题效果: 鉴于国际棋盘两点,寻求同意的操作,是什么操作的最小数量,在操作过程中输出. 解题思路: 水题一个,见代码. 以下是代码: #include <set> #include <map> #include <queue> #include <math.h> #include <vector> #include <string> #include <stdio.h> #include <string.h>…