题意: M行N列矩阵, 'Y'表示开始位置, 'T'表示目标位置, 从开始位置到目标位置至少需要走多少步,其中, 'S', 'R'表示不能走, 'B' 花费为2, 'E'花费为1. 思路:纯 BFS. 很久没写BFS, 刚写居然不知道从何下手... #include <iostream> #include <queue> using namespace std; struct postion { int i, j; postion operator+(postion p) { po…

[思路] 题目中的“可以沿直线发射打破砖墙”可能会迷惑到很多人,实际上可以等价理解为“通过砖墙的时间为2个单位”,这样题目就迎刃而解了.第一次碰到时可能不能很好把握,第二次基本就可以当作水题了. [错误点] 1.不能用裸的bfs.广搜的实际思想是将到达时间最短的放在队首,这样首次到达终点即为时间的最小值.通过砖墙的时间为两个单位,通过砖墙后可能不是时间最小值.用优先队列可以解决这一问题. 2.c++中memset初始化为memset(vis,0,sizeof(vis)),很多人可能会写成mems…

来源poj2312 Many of us had played the game "Battle city" in our childhood, and some people (like me) even often play it on computer now. What we are discussing is a simple edition of this game. Given a map that consists of empty spaces, rivers, st…

Battle City Many of us had played the game "Battle city" in our childhood, and some people (like me) even often play it on computer now. What we are discussing is a simple edition of this game. Given a map that consists of empty spaces, rivers,…

Many of us had played the game "Battle city" in our childhood, and some people (like me) even often play it on computer now. What we are discussing is a simple edition of this game. Given a map that consists of empty spaces, rivers, steel walls…

题目:1013 Battle Over Cities 思路:城市数也就1000, 对于每次询问暴力bfs一下看一下有多少连通块就行了.答案就是联通块数减一. #include<stdio.h> #include<stdlib.h> #include<map> #include<set> #include<iostream> #include<cstring> #include<algorithm> #include<…

  Battle City Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7579   Accepted: 2544 Description Many of us had played the game "Battle city" in our childhood, and some people (like me) even often play it on computer now. What we are…

Battle City Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9885   Accepted: 3285 Description Many of us had played the game "Battle city" in our childhood, and some people (like me) even often play it on computer now. What we are d…

Many of us had played the game "Battle city" in our childhood, and some people (like me) even often play it on computer now. What we are discussing is a simple edition of this game. Given a map that consists of empty spaces, rivers, steel walls…

题意:给你一个地图,’+’代表十字路口,‘-’‘|’表示街道,‘.’表示建筑物,‘s’,’E’ 起点和终点.输出从起点到终点的的 最短路径(包括方向和沿该方向的经过的十字路口数) 分析:ans[i][j],起点到(i,j)点的最短路径,bfs求出所有,再从终点回推到起点得最短路径. #include <map> #include <set> #include <list> #include <cmath> #include <queue> #in…