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…

题目链接:http://poj.org/problem?id=2312 题目大意:给出一个n*m的矩阵,其中Y是起点,T是终点,B和E可以走,S和R不可以走,要注意的是走B需要2分钟,走E需要一分钟.最后求解Y--->T的最短时间!! 看到这题首先想到广搜来找最短时间,但是这里可以对B和E进行处理,方便计算~ #include <iostream> #include <cstdio> #include <queue> #include <cstring>…

  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…

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…

题目连接 http://poj.org/problem?id=1840 Battle City 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 discussing is a simple edition of this game. Giv…

普通BFS:每个状态只访问一次,第一次入队时即为该状态对应的最优解. 优先队列BFS:每个状态可能被更新多次,入队多次,但是只会扩展一次,每次出队时即为改状态对应的最优解. 且对于优先队列BFS来说,每次存入队列的不光是像普通BFS的状态,还有当前状态对应的代价,并且是依据最小代价进行扩展.每次状态被更近之后,将其入队. 对于本题来说 状态选取:当前所在城市,当前油量 代价函数:当前状态所对应的最小花费 代码如下: #include <cstdio> #include <iostream…

Battle City Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7208   Accepted: 2427 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…