Rescue Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 14   Accepted Submission(s) : 7 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Angel was caught by the MOLIGPY…

Rescue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19985    Accepted Submission(s): 7110 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is d…

题意:X代表卫兵,a代表终点,r代表起始点,.代表路,#代表墙,走过.要花费一秒,走过x要花费2秒,求从起点到终点的最少时间. 析:一看到样例就知道是BFS了吧,很明显是最短路径问题,不过又加了一个条件——时间,所以我们用优先队列去优先获取时间短的路径,总体实现起来没有太大难度. 代码如下: #include <iostream> #include <cstdio> #include <vector> #include <set> #include <…

War chess is hh's favorite game: In this game, there is an N * M battle map, and every player has his own Moving Val (MV). In each round, every player can move in four directions as long as he has enough MV. To simplify the problem, you are given you…

题目传送门 题意:从r走到a,遇到x多走一步,问最小走到a的步数 分析:因为r有多个,反过来想从a走到某个r的最小步数,简单的BFS.我对这题有特殊的感情,去年刚来集训队时肉鸽推荐了这题,当时什么都不会,看个数组模拟队列的BFS看的头晕,现在看起来也不过如此,额,当年开始是从r走到a的,因为数据巨弱才过的,应该要用到优先队列. /************************************************ * Author :Running_Time * Created Ti…

Rescue Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21510 Accepted Submission(s): 7671 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is descri…

Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10541    Accepted Submission(s): 3205Special Judge Problem Description The Princess has been abducted by the BEelzebub…

题目 用优先队列优化普通的广搜就可以过了. #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #include<queue> struct pq { int x,y,val; friend bool operator < (pq a,pq b) { return a.val > b.val; } }; priority_queue <p…

题目大意: 在n个点 m条边的无向图中 需要运送X单位牛奶 每条边有隐患L和容量C 则这条边上花费时间为 L+X/C 求从点1到点n的最小花费 优先队列维护 L+X/C 最小 广搜到点n #include <bits/stdc++.h> using namespace std; #define LL long long #define INF 0x3f3f3f3f #define mem(i,j) memset(i,j,sizeof(i)) #define inc(i,l,r) for(int…

题目链接:hdu 1242 这题也是迷宫类搜索,题意说的是 'a' 表示被拯救的人,'r' 表示搜救者(注意可能有多个),'.' 表示道路(耗费一单位时间通过),'#' 表示墙壁,'x' 代表警卫(耗费两个单位时间通过),然后求出 'r' 能找到 'a' 的最短时间,找不到输出 "…………"(竟然在这里也 wa 了一发 -.-||).很明显是广搜了,因为 'r' 可能有多个,所以我们反过来从 'a' 开始搜,每次搜到 'r' 都更新最小时间值(很重要的一个转换!).可是这题因为通过 '…