Walk Out
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3210 Accepted Submission(s): 647
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.
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.
0
unless the answer itself is 0
(in this case, print 0
instead).
首先很容易想到位数越少越小，所以说肯定选择向下或者向右的走向到终点(即最短路径为忧)
其次如果一开始(1,1)为0的话，如果有一段连续的0路径，可以选择先绕到离终点最近的0，这样前面全是前导0，对答案没有影响
所以说策略是先找到一段连续的0距终点最近，然后再在每层寻找最小的数字(这里说的层和距离都是斜过来的)
千万不能用dfs找每层的0....数据卡了这个，直接每次递推寻找最小值然后标记就好了(哭死了，当时因为这个超时没过)
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define N 1005 int n,m;
int tx[4]={0,0,1,-1},ty[4]={1,-1,0,0};
char graph[N][N];
bool used[N][N];
int xx[1100000], yy[1100000]; void Bfsimilar(){
int i,j,k;
memset(used,false,sizeof(used));
used[1][1] = true;
int q=1,h=1;
xx[q]=yy[q]=1;
for (; q<=h ; q++) //递归形dfs拿时间换空间
if(graph[xx[q]][yy[q]]=='0'){
for(i=0;i<4;i++){
int X=xx[q]+tx[i], Y=yy[q]+ty[i];
if(X>0 && X<=n && Y>0 && Y<=m && !used[X][Y]){ //找到最近的1
h++;
xx[h]=X;
yy[h]=Y;
used[X][Y]=true;
}
}
}
if(used[n][m] && graph[n][m]=='0') { //处理一直是0的情况
printf("0\n");
return;
}
int ma=0;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
if(used[i][j])
ma=max(ma,i+j); //找到最近的0
printf("1");
//printf("%d\n",ma);
for(i=ma;i<n+m;i++){
char mi='1';
int temp1=max(1,i-m);
int temp2=min(n,i-1);
for(j=temp1;j<=temp2;j++)
if(used[j][i-j]){
mi=min(mi, graph[j+1][i-j]);
mi=min(mi, graph[j][i-j+1]);
}
printf("%c",mi);
for(j=temp1;j<=temp2;j++)
if(used[j][i-j]){
if(graph[j+1][i-j]==mi) used[j+1][i-j] =true;
if(graph[j][i-j+1]==mi) used[j][i-j+1] =true;
}
}
printf("\n");
} int main() {
//freopen("in.txt", "r", stdin);
int T;
int i,j,k;
scanf("%d",&T);
while(T--){
scanf("%d %d",&n,&m);
for(i=1;i<=n;i++)
scanf("%s",graph[i]+1);
for(i=0;i<=n+1;i++)
graph[i][0]='2',graph[i][m+1]='2'; //将边界处理为2，方便之后的处理
for(i=0;i<= m+1;i++)
graph[0][i]='2',graph[n+1][i]='2';
Bfsimilar();
}
return 0;
}