HOJ---12363 Robots on a grid [DP+BFS()]

Robots on a grid
Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65535KB
Total submit users: 30, Accepted users: 24
Problem 12363 : No special judgement
Problem description

You have recently made a grid traversing robot that can find its way from the top left corner of a grid to the bottom right corner. However, you had forgotten all your AI programming skills, so you only programmed your robot to go rightwards and downwards (that's after all where the goal is). You have placed your robot on a grid with some obstacles, and you sit and observe. However, after a while you get tired of observing it getting stuck, and ask yourself "How many paths are there from the start position to the goal position?", and "If there are none, could the robot have made it to the goal if it could walk upwards and leftwards?"
So you decide to write a program that, given a grid of size nn with some obstacles marked on it where the robot cannot walk, counts the different ways the robot could go from the top left corner s to the bottom right t, and if none,tests if it were possible if it could walk up and left as well. However, your program does not handle very large numbers, so the answer should be given modulo 231-1.
Input
On the first line is one integer, 1 <= n <= 1000. Then follows n lines, each with n characters,where each character is one of '.' and '#', where '.' is to be interpreted as a walkable tile and '#' as a non-walkable tile. There will never be a wall at s, and there will never be a wall at t.
Output
Output one line with the number of di erent paths starting in s and ending in t (modulo 231-1) or THE GAME IS A LIE if you cannot go from s to t going only rightwards and downwards but you can if you are allowed to go left and up as well, or INCONCEIVABLE if there simply is no path from s to t.
Sample Input 
5
.....
#..#.
#..#.
...#.
.....
7
......#
####...
.#.....
.#...#.
.#.....
.#..###
.#.....
Sample Output 
6
THE GAME IS A LIE
Problem Source

NCPC 2011

 

 

 

 

要用bfs

code ：

  1 /*
  2 5
  3 .....
  4 #..#.
  5 #..#.
  6 ...#.
  7 .....
  8 7
  9 ......#
 10 ####...
 11 .#.....
 12 .#...#.
 13 .#.....
 14 .#..###
 15 .#.....
 16 7
 17 ......#
 18 ####...
 19 .#.....
 20 .#...#.
 21 .#.....
 22 .#..###
 23 .#...#.
 24 5
 25 .....
 26 #..#.
 27 #..#.
 28 ...#.
 29 ...#.
 30 5
 31 .....
 32 #..#.
 33 #..#.
 34 ...##
 35 ...#.
 36 5
 37 .....
 38 #..#.
 39 #..#.
 40 .....
 41 ...#.
 42 */
 43 #include <iostream>   
 44 #include <iomanip>   
 45 #include <fstream>   
 46 #include <sstream>   
 47 #include <algorithm>   
 48 #include <string>   
 49 #include <set>   
 50 #include <utility>   
 51 #include <queue>   
 52 #include <stack>   
 53 #include <list>   
 54 #include <vector>   
 55 #include <cstdio>   
 56 #include <cstdlib>   
 57 #include <cstring>   
 58 #include <cmath>   
 59 #include <ctime>   
 60 #include <ctype.h> 
 61 using namespace std;
 62 
 63 char map[1010][1010];
 64 __int64 dp[1010][1010];
 65 int vst[1010][1010];
 66 bool flag;
 67 int n;
 68 
 69 struct node
 70 {
 71     int x,y;
 72 }Node;
 73 
 74 int dir_bfs[4][2]={
 75     0,1,
 76     1,0,
 77     0,-1,
 78     -1,0
 79 };
 80 
 81 bool check_bfs(int x,int y)
 82 {
 83     if(x>=0&&x<n&&y>=0&&y<n&&map[x][y]!='#'&&!vst[x][y])
 84         return true;
 85     else
 86         return false;
 87 }
 88 
 89 void bfs()
 90 {
 91     int i;
 92     node pre,last;
 93     pre.x=0;
 94     pre.y=0;
 95     queue<node>Que;
 96     Que.push(pre);
 97     while(!Que.empty())
 98     {
 99         pre=Que.front();
100         Que.pop();
101         if(pre.x==n-1&&pre.y==n-1)
102         {
103             flag=true;
104             return;
105         }
106         for(i=0;i<4;i++)
107         {
108             last.x=pre.x+dir_bfs[i][0];
109             last.y=pre.y+dir_bfs[i][1];
110             if(check_bfs(last.x,last.y))
111             {
112                 Que.push(last);
113                 vst[last.x][last.y]=1;
114             }
115         }
116     }
117 }
118 
119 int main()
120 {
121     int i,j;
122     while(~scanf("%d",&n))
123     {
124         getchar();
125         memset(dp,0,sizeof(dp));
126         for(i=0;i<n;i++)
127             gets(map[i]);
128         for(i=0;i<n;i++)
129         {
130             if(map[i][0]=='#')
131                 break;
132             dp[i][0]=1;
133         }
134         for(i=0;i<n;i++)
135         {
136             if(map[0][i]=='#')
137                 break;
138             dp[0][i]=1;
139         }
140         for(i=1;i<n;i++)
141         {
142             for(j=1;j<n;j++)
143             {
144                 if(map[i][j]!='#')
145                     dp[i][j]=(dp[i][j-1]+dp[i-1][j])%INT_MAX;
146             }
147         }
148         if(dp[n-1][n-1])
149             printf("%I64d\n",dp[n-1][n-1]%INT_MAX);
150         else
151         {
152             flag=false;
153             memset(vst,0,sizeof(vst));
154             bfs();
155             if(flag)
156                 printf("THE GAME IS A LIE\n");
157             else
158                 printf("INCONCEIVABLE\n");
159         }
160     }
161     return 0;
162 }

 

posted @ 2012-07-29 10:58  max_xbw  阅读(326)  评论(0)  编辑  收藏  举报