题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2822 Dogs Description Prairie dog comes again! Someday one little prairie dog Tim wants to visit one of his friends on the farmland, but he is as lazy as his friend (who required Tim to come to his place…

http://acm.hdu.edu.cn/showproblem.php?pid=2822 给定起点和终点,问从起点到终点需要挖几次只有从# 到 .或者从. 到  . 才需要挖一次. #include <cstdio> #include <queue> #include <cstring> using namespace std; ; int n,m; int sx,sy,ex,ey; char maze[maxn][maxn]; int vis[maxn][maxn…

题目链接:hdu2822 会优先队列话这题很容易AC.... #include<stdio.h> #include<string.h> #include<queue> #include<algorithm> #define N 1005 using namespace std; char map[N][N]; int v[N][N],d[4][2] = { {-1,0},{1,0},{0,-1},{0,1} }; int begin_x,begin_y,en…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2822 题目大意:X消耗0,.消耗1, 求起点到终点最短消耗 解题思路: 每层BFS的结点,优先级不同,应该先搜cost小的.直接退化为最短路问题. 优先队列优化. 卡输入姿势.如果O(n^2)逐个读的话会T掉.要用字符串读一行. #include "cstdio" #include "queue" #include "cstring" using…

Dogs' Candies Time Limit: 30000/30000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 1701 Accepted Submission(s): 404 Problem Description Far far away, there live a lot of dogs in the forest. Unlike other dogs, thos…

bfs. /* 2822 */ #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <queue> using namespace std; #define MAXN 1005 #define INF 0xffffff typedef struct node_t { int d, x, y; node_t() {} node_t(…

Dogs Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2208    Accepted Submission(s): 838 Problem Description Prairie dog comes again! Someday one little prairie dog Tim wants to visit one of his…

周六周末组队训练赛. Dogs' Candies Time Limit: 30000/30000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 3920    Accepted Submission(s): 941 Problem Description Far far away, there live a lot of dogs in the forest. Unlike…

首先 在此哀悼...  为我逝去的时间哀悼...  每一步都确定再去写下一步吧...日狗 不过还是有点收获的..  对优先队列的使用 有了进一步的理解 先上代码 #include<iostream>#include<cstdio>#include<queue>#include<string.h>using namespace std;int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};char mapp[1001][1001]…

一.预备知识 \(tD/eD\) 问题:状态 t 维,决策 e 维.时间复杂度\(O(n^{e+t})\). 四边形不等式: 称代价函数 w 满足凸四边形不等式,当:\(w(a,c)+w(b,d)\le w(b,c)+w(a,d),\ a < b < c < d\) 如下所示,区间1.2对应的 w 之和 ≤ 3.4之和 \[ \underbrace {\overbrace {a \to \underbrace{b \to c}_3}^1 \to d }_4 \llap{\overbrac…