Walk Out
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1167 Accepted Submission(s): 216
Problem Description
In an n∗m maze, the right-bottom corner is the exit (position (n,m) is the exit). In every position of this maze, there is either a 0 or a 1 written on it.
An explorer gets lost in this grid. His position now is (1,1), and he wants to go to the exit. Since to arrive at the exit is easy for him, he wants to do something more difficult. At first, he'll write down the number on position (1,1). Every time, he could make a move to one adjacent position (two positions are adjacent if and only if they share an edge). While walking, he will write down the number on the position he's on to the end of his number. When finished, he will get a binary number. Please determine the minimum value of this number in binary system.
An explorer gets lost in this grid. His position now is (1,1), and he wants to go to the exit. Since to arrive at the exit is easy for him, he wants to do something more difficult. At first, he'll write down the number on position (1,1). Every time, he could make a move to one adjacent position (two positions are adjacent if and only if they share an edge). While walking, he will write down the number on the position he's on to the end of his number. When finished, he will get a binary number. Please determine the minimum value of this number in binary system.
Input
The first line of the input is a single integer T (T=10), indicating the number of testcases.
For each testcase, the first line contains two integers n and m (1≤n,m≤1000). The i-th line of the next n lines contains one 01 string of length m, which represents i-th row of the maze.
For each testcase, the first line contains two integers n and m (1≤n,m≤1000). The i-th line of the next n lines contains one 01 string of length m, which represents i-th row of the maze.
Output
For each testcase, print the answer in binary system. Please eliminate all the preceding 0 unless the answer itself is 0 (in this case, print 0 instead).
Sample Input
2
2 2
11
11
3 3
111
111
111
Sample Output
111
101
Author
XJZX
Source
/** 题意：给一个n*m的01矩阵 然后要求从(0,0) 走到(n-1,m-1) 问走到的最小的串 做法：bfs + 贪心先找离(n-1,m-1)，最近的1的位置，就是找所有的 前缀0，然后从最近的1开始搜，只需要搜索当前位置的左和下 然后直至(n-1,m-1) **/ #include <iostream> #include <algorithm> #include <cmath> #include <stdio.h> #include <string.h> #include <queue> using namespace std; #define maxn 1100 int vis[maxn][maxn]; char ch[maxn][maxn]; int n, m; int dx[4][2] = {1, 0, 0, 1, -1, 0, 0, -1}; int sx, sy; struct Node { int x; int y; Node() {} }; int check(int x, int y) { if(x >= 0 && x < n && y >= 0 && y < m) { return 1; } return 0; } void bfs(int x, int y) { Node tmp, now, temp; queue<Node>que; vis[x][y] = 1; temp.x = x; temp.y = y; que.push(temp); while(!que.empty()) { now = que.front(); que.pop(); for(int i = 0; i < 4; i++) { tmp.x = now.x + dx[i][0]; tmp.y = now.y + dx[i][1]; if(check(tmp.x, tmp.y) && vis[tmp.x][tmp.y] == 0) { vis[tmp.x][tmp.y] = 1; if(ch[tmp.x][tmp.y] == '0') { que.push(tmp); } if(tmp.x + tmp.y > sx + sy) { sx = tmp.x; sy = tmp.y; } } } } } void bfs1() { printf("1"); bool isok = false; bool isok1 = false; for(int i = sx + sy; i < n + m - 2; i++) { isok = false; for(int j = 0; j <= i; j++) { int x = j; int y = i - j; if(check(x, y) == 0 || vis[x][y] == 0) { continue; } if(isok1 && ch[x][y] == '1') { continue; } for(int p = 0; p < 2; p++) { int tx = x + dx[p][0]; int ty = y + dx[p][1]; if(check(tx, ty) == 0) { continue; } vis[tx][ty] = 1; if(ch[tx][ty] == '0') { isok = true; } } } isok1 = isok; if(isok) { printf("0"); } else { printf("1"); } } printf(" "); } int main() { int T; scanf("%d", &T); while(T--) { scanf("%d %d", &n, &m); memset(vis, 0, sizeof(vis)); for(int i = 0; i < n; i++) { scanf("%s", ch[i]); } sx = sy = 0; vis[0][0] = 1; if(ch[0][0] == '0') { bfs(0, 0); } //cout << sx << " " << sy << endl; if(ch[sx][sy] == '0') { printf("0 "); } else { bfs1(); } } return 0; }