Walk Out
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, 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
001
111
101
Sample Output
111
101
Author
XJZX
Source
Recommend
本题思路：
1.先判断第一个点是不是0，如果是0先把所有的0都走一遍，找到哈曼顿距离最小的点（可能会有几个）。
这个过程可以用DFS也可以BFS（建议BFS,因为DFS会爆栈，必须自己把栈开导最大，后面会说明）
2.如果第一点不是0，直接从第一个点开始搜，只搜下和右两个方向，如果这两个方向有两个0，输出0，把两个0都加入队列；如果只有一个0，输出0，只把0那个点加入队列；如果是两个1，也把两个点都加入队列。
3.如果第一个点是0，再把这个0的右边的点和下边的点（超边界的不算）都加入队列开始用2的方法搜。
开始用DFS搜
1 #pragma comment(linker, "/STACK:10240000000000,10240000000000")//这行代码不加就会STACK_OVERFLOW 2 #include<queue> 3 #include<math.h> 4 #include<stdio.h> 5 #include<string.h> 6 #include<iostream> 7 #include<algorithm> 8 using namespace std; 9 #define N 1234 10 struct point 11 { 12 int x,y,d; 13 }st[N*2]; 14 15 int dx[]={1,0,-1,0}; 16 int dy[]={0,1,0,-1}; 17 int n,m,k,dis,flag; 18 char mat[N][N]; 19 bool vis[N][N]; 20 21 void dfs(int x,int y) 22 { 23 if(x<1||x>n||y<1||y>m)return; 24 if(mat[x][y]=='1')return; 25 if(vis[x][y]==1)return; 26 vis[x][y]=1; 27 if(dis<x+y) 28 { 29 k=0; 30 dis=x+y; 31 st[k].x=x; 32 st[k++].y=y; 33 } 34 else if(dis==x+y) 35 { 36 st[k].x=x; 37 st[k++].y=y; 38 } 39 for(int i=0;i<4;i++) 40 dfs(x+dx[i],y+dy[i]); 41 } 42 void bfs() 43 { 44 memset(vis,0,sizeof(vis)); 45 queue<point>q1; 46 queue<point>q2; 47 for(int i=0;i<k;i++) 48 { 49 if(mat[st[i].x][st[i].y]=='0') 50 { 51 if(st[i].x==n&&st[i].y==m){printf("0");return;} 52 point a1=st[i],a2=st[i]; 53 a1.x++;a2.y++; 54 if(a1.x<=n) 55 q1.push(a1),vis[a1.x][a1.y]=1; 56 if(a2.y<=m) 57 q1.push(a2),vis[a2.x][a2.y]=1; 58 } 59 else 60 q1.push(st[i]),vis[st[i].x][st[i].y]=1; 61 } 62 printf("1"); 63 if(vis[n][m])return; 64 while(1) 65 { 66 flag=1; 67 while(!q1.empty()) 68 { 69 point cur=q1.front(); 70 q1.pop(); 71 for(int i=0;i<2;i++) 72 { 73 point next=cur; 74 next.x+=dx[i],next.y+=dy[i]; 75 if(vis[next.x][next.y] || next.x<1 || next.x>n || next.y<1 || next.y>m)continue; 76 if(mat[next.x][next.y] == '0') 77 flag = 0; 78 q2.push(next); 79 vis[next.x][next.y]=1; 80 } 81 } 82 printf("%d",flag); 83 if(vis[n][m])return; 84 85 while(!q2.empty()) 86 { 87 point cur=q2.front(); 88 q2.pop(); 89 if(flag==1) 90 q1.push(cur); 91 else if(flag==0 && mat[cur.x][cur.y]=='0') 92 q1.push(cur); 93 } 94 } 95 } 96 97 int main() 98 { 99 int t;cin>>t; 100 while(t--) 101 { 102 memset(vis,0,sizeof(vis)); 103 dis=0; 104 scanf("%d%d",&n,&m); 105 for(int i=1;i<=n;i++) 106 scanf("%s",mat[i]+1); 107 108 if(mat[1][1]=='1') 109 st[0].x=1,st[0].y=1,k=1; 110 else 111 dfs(1,1); 112 bfs(); 113 cout<<endl; 114 } 115 return 0; 116 } 117 118 //几组比较好的数据 119 120 /* 121 5 122 2 2 123 01 124 11 125 2 2 126 00 127 11 128 2 2 129 00 130 00 131 3 3 132 000 133 110 134 110 135 3 3 136 000 137 110 138 111 139 140 141 */
开始用BFS搜（推荐）
1 #include<queue> 2 #include<math.h> 3 #include<stdio.h> 4 #include<string.h> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 #define N 1234 9 struct point 10 { 11 int x,y; 12 }st[N*2]; 13 int dx[]={1,0,-1,0}; 14 int dy[]={0,1,0,-1}; 15 int n,m,k,dis; 16 char mat[N][N]; 17 bool vis[N][N]; 18 19 void bfs1() 20 { 21 memset(vis,0,sizeof(vis)); 22 queue<point>q; 23 point first; 24 first.x=first.y=1; 25 q.push(first);vis[1][1]=1; 26 st[0].x=st[0].y=1; 27 k=1; 28 while(!q.empty()) 29 { 30 point cur=q.front(); 31 q.pop(); 32 for(int i=0;i<4;i++) 33 { 34 point next=cur; 35 next.x+=dx[i],next.y+=dy[i]; 36 if(next.x<1||next.x>n||next.y<1||next.y>m)continue; 37 if(vis[next.x][next.y] || mat[next.x][next.y]=='1')continue; 38 q.push(next);vis[next.x][next.y]=1; 39 if(dis<next.x+next.y) 40 { 41 k=0; 42 dis=next.x+next.y; 43 st[k].x=next.x; 44 st[k++].y=next.y; 45 } 46 else if(dis==next.x+next.y) 47 { 48 st[k].x=next.x; 49 st[k++].y=next.y; 50 } 51 } 52 } 53 } 54 void bfs() 55 { 56 memset(vis,0,sizeof(vis)); 57 queue<point>q1; 58 queue<point>q2; 59 for(int i=0;i<k;i++) 60 { 61 if(mat[st[i].x][st[i].y]=='0') 62 { 63 if(st[i].x==n&&st[i].y==m){printf("0");return;} 64 point a1=st[i],a2=st[i]; 65 a1.x++;a2.y++; 66 if(a1.x<=n) 67 q1.push(a1),vis[a1.x][a1.y]=1; 68 if(a2.y<=m) 69 q1.push(a2),vis[a2.x][a2.y]=1; 70 } 71 else 72 q1.push(st[i]),vis[st[i].x][st[i].y]=1; 73 } 74 printf("1"); 75 if(vis[n][m])return; 76 while(1) 77 { 78 int flag=1; 79 while(!q1.empty()) 80 { 81 point cur=q1.front(); 82 q1.pop(); 83 for(int i=0;i<2;i++) 84 { 85 point next=cur; 86 next.x+=dx[i],next.y+=dy[i]; 87 if(vis[next.x][next.y] || next.x<1 || next.x>n || next.y<1 || next.y>m)continue; 88 if(mat[next.x][next.y] == '0') 89 flag = 0; 90 q2.push(next); 91 vis[next.x][next.y]=1; 92 } 93 } 94 printf("%d",flag); 95 if(vis[n][m])return; 96 97 while(!q2.empty()) 98 { 99 point cur=q2.front(); 100 q2.pop(); 101 if(flag==1) 102 q1.push(cur); 103 else if(flag==0 && mat[cur.x][cur.y]=='0') 104 q1.push(cur); 105 } 106 } 107 } 108 109 int main() 110 { 111 int t;cin>>t; 112 while(t--) 113 { 114 dis=0; 115 scanf("%d%d",&n,&m); 116 for(int i=1;i<=n;i++) 117 scanf("%s",mat[i]+1); 118 119 if(mat[1][1]=='1') 120 st[0].x=1,st[0].y=1,k=1; 121 else 122 bfs1(); 123 bfs(); 124 cout<<endl; 125 } 126 return 0; 127 }
其他：
1输图的时候不要%c输入，用%s输入，速度会快很多，这题如果用%c输入会TLE,(花了一下午时间找为什么TLE，最后发现居然是因为输图方式。） 所以以后都要用：
for(int i=0;i<n;i++) scanf("%s",mat[i]; or for(int i=1;i<=n;i++) scanf("%s",mat[i]+1);
2 dfs是很容易爆栈的，这题就是我开始写的用dfs的就爆栈了，这时候有一个处理办法：在代码最前面加：#pragma comment(linker, "/STACK:10240000000000,10240000000000") 这句意思是自己开一个非常大的栈，STACK:后面那数字好像已经是能开的最大的了。
此题中加入这一行本来的Runtime Error（STACK_OVERFLOW)就会变成 Accepted！
但是好像正规比赛不允许使用这种方式。