题意：在n×m的地图上，0表示墙，1表示空地，2表示人，3表示目的地，4表示有定时炸弹重启器。定时炸弹的时间是6，人走一步所需要的时间是1。每次可以上、下、左、右移动一格。当人走到4时如果炸弹的时间不是0，可以...
题意：在n×m的地图上，0表示墙，1表示空地，2表示人，3表示目的地，4表示有定时炸弹重启器。定时炸弹的时间是6，人走一步所需要的时间是1。每次可以上、下、左、右移动一格。当人走到4时如果炸弹的时间不是0，可以...
DescriptionIgnatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes....
https://cn.vjudge.net/problem/15705/origin dfs的一道好题，注意减枝 题意：要求从第一个星球遍历访问其他星球，然后输出遍历每个星球所用时间的总和，就酱 #include&...algorithm...
#include using namespace std; int map[10][10]; int _time[10][10]; int step[10][10]; int T,N,M; int mins; int sx,sy; int dirx[4] = {-1,0,1,0}; int diry[4] = {0,1,0,-1};...void DFS(int x,in
Nightmare Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 1 Accepted Submission(s) : 1 ...Font: Times New Roman | Verdana | Georgia ...
Problem Description Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb ...
Nightmare Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): ...
#include<iostream> #include<queue> #include<string> using namespace std; typedef struct point { int x,y; int step;...int map[10]...
题目大意： 0为墙1为路2为起点3为终点4为炸弹 走到任意一个炸弹都可以将所有炸弹重置倒计时6minutes 每走一个位置需要1minutes 问从2到3需要的最少时间 ...处理走到同一个炸弹位置方法：第一次走到炸弹的位置时，...
Nightmare Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11206 Accepted Submission(s): 5457 Problem Description Ignatius had a
Nightmare Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10281 Accepted Submission(s): 5015 Problem Description ...Ignatius had a
本题目由于普通的个子可以重复走（在样例三中可以看出），所以不能通过标记位限制只能走一次，但是需要进行一定的限制防止无限的循环。实际上，为4的格子只需要走一次，重复走只会使得步数变的更多，并不会增加到达...
source page：http://acm.hdu.edu.cn/showproblem.php?pid=1072 Nightmare Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 80
#include #include #include #include #include #include #include #include #include #include using namespace std; #include #include #include #include #include #include ... int
Nightmare Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9400 Accepted Submission(s): 4533 Problem Description ...Ignatius had a n
Nightmare Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9120 Accepted Submission(s): 4389 Problem Description ...Ignatius had a night
hdu1072题解 Nightmare Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14457 Accepted Submission(s): 7027 Problem Description Ignatius had a ...
hdu1072 nightmare ……跟hdu1242有些不同 走过的路可以返回 所以不要标记，换炸弹的倒计时限制队列的增长 #include #include #include using namespace std; int mp[10][10],m,n,Sx,Sy,Dx,Dy,flag...
#include<iostream> #include<algorithm> #include<cstdio> #include<string> #include<vector> #include<string.h> #include<map> #include<...#de
#include #include #include using namespace std; int a[10][10]; int state_time[10][10]; int f[4][2]={0,1,0,-1,1,0,-1,0}; struct state { int x,y; int time; int count; }; int main() ... int
题目链接 Nightmare 炸弹爆炸的限时是6s,一步减掉1s，在'4'方块上炸弹爆炸时间重置为6s 但是走到'4'和终点时的剩余时间必须大于0 ...注意每点可以重复走过，必须剪枝，否则就停不下来了 ...如果第二次走过‘1’时...