思路 : 优先队列 每次都取最小的时间,遇到了终点直接就输出 #include<iostream> #include<queue> #include<cstring> using namespace std; ; int n,m,sa,ea,sr,er; char arr[N][N]; }; struct stu{ int a,b; int time; bool friend operator <(const stu &a,const stu &b…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1242 题目描述: Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison…
Rescue Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20938 Accepted Submission(s): 7486 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is d…
Rescue Time Limit: 2 Seconds Memory Limit: 65536 KB Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want…
直接把Angle的位置作为起点,广度优先搜索即可,这题不是步数最少,而是time最少,就把以time作为衡量标准,加入优先队列,队首就是当前time最少的.遇到Angle的朋友就退出.只需15ms AC代码: #include<cstdio> #include<cstring> #include<queue> using namespace std; const int maxn=202; int d[maxn][maxn]; int n,m; char G[maxn]…
