hdu 5335 Walk Out 搜索+贪心
Walk Out
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 141 Accepted Submission(s): 17
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 3 001 111 101
Sample Output
111 101
Source
2015 Multi-University Training Contest 4
求从1,1到n，n的路径中，得到的二进制数最小是什么
方法：
如果1,1是0那么搜索出所有可达的0号点，前缀0为0，所以可以从这些位置出发。
然后有了一个1,。
那么只要路径最短的，同时得到的二进制数最小的。
从所有可出发的点出发，做bfs，计算到n，n的最短路。
然后从n，n做bfs看，那些位置是在最短路上的。
然后从所有可出发的点出发，bfs。如果存在能到的0点。则从这些点出发，否则从到的1点出发。
这样贪心做即可，直到到n，n点
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define maxn 1001
char ma[maxn][maxn];
int check[maxn][maxn];
int dep[maxn][maxn];
struct Node{
int x,y;
Node(int _x=0,int _y=0):x(_x),y(_y){}
};
queue<Node> que;
vector<int> ans;
vector<Node> one;
vector<Node> zero;
int n,m;
void dfs(int x,int y){
if(check[x][y] || x == n || y == m || x < 0 || y < 0 || ma[x][y] == '1') return;
que.push(Node(x,y));
check[x][y] = 1;
dfs(x+1,y);
dfs(x,y+1);
dfs(x-1,y);
dfs(x,y-1);
}
void work(){
while(que.size() > 0)
que.pop();
if(ma[0][0] == '1'){
check[0][0] = 1;
que.push(Node(0,0));
}
else {
dfs(0,0);
}
}
void bfs(){
Node a;
while(que.size() > 0){
a = que.front();
que.pop();
if(a.x + 1 < n && a.y < m && check[a.x+1][a.y] == 0){
que.push(Node(a.x+1,a.y));
check[a.x+1][a.y] = 1;
dep[a.x+1][a.y] = dep[a.x][a.y] + 1;
}
if(a.x < n && a.y + 1 < m && check[a.x][a.y+1] == 0){
que.push(Node(a.x,a.y+1));
check[a.x][a.y+1] = 1;
dep[a.x][a.y+1] = dep[a.x][a.y] + 1;
}
}
}
int mark[maxn][maxn];
void nidfs(int x,int y){
mark[x][y] = 1;
if(x-1>=0 && y >= 0 && mark[x-1][y] == 0 && dep[x-1][y] == dep[x][y] - 1 && dep[x-1][y] > 0)
nidfs(x-1,y);
if(x >= 0 && y-1>=0 && mark[x][y-1] == 0 && dep[x][y-1] == dep[x][y] - 1 && dep[x][y-1] > 0)
nidfs(x,y-1);
}
void getans(){
ans.clear();
Node a;
while(1){
one.clear();
zero.clear();
while(que.size() > 0){
a = que.front();
que.pop();
if(a.x == n -1 && a.y == m -1 ) return ;
if(a.x + 1 < n && a.y < m && check[a.x+1][a.y] == 0&& dep[a.x+1][a.y] == dep[a.x][a.y] + 1 && mark[a.x+1][a.y]){
if(ma[a.x+1][a.y] == '1') one.push_back(Node(a.x+1,a.y));
else zero.push_back(Node(a.x+1,a.y));
check[a.x+1][a.y] = 1;
}
if(a.x < n && a.y + 1< m && check[a.x][a.y+1] == 0 && dep[a.x][a.y+1] == dep[a.x][a.y] + 1 && mark[a.x][a.y+1]){
if(ma[a.x][a.y+1] == '1') one.push_back(Node(a.x,a.y+1));
else zero.push_back(Node(a.x,a.y+1));
check[a.x][a.y+1] = 1;
}
}
if(zero.size() > 0){
ans.push_back(0);
for(int i = 0;i < zero.size();i++)
que.push(zero[i]);
}
else {
ans.push_back(1);
for(int i = 0;i < one.size();i++)
que.push(one[i]);
}
}
}
int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for(int i = 0;i < n; i++){
scanf("%s",ma[i]);
}
memset(check,0,sizeof(check));
work();
memset(dep,0,sizeof(dep));
bfs();
// for(int i = 0;i < n; i++){
// for(int j = 0;j < m; j++){
// printf("%d ",dep[i][j]);
// }cout<<endl;
// }
// cout<<endl;
memset(mark,0,sizeof(mark));
nidfs(n-1,m-1);
memset(check,0,sizeof(check));
work();
getans();
// for(int i = 0;i < n; i++){
// for(int j = 0;j < m; j++){
// printf("%d ",dep[i][j]);
// }cout<<endl;
// }
if(ans.size() == 0 && ma[0][0] =='0'){
printf("0\n");
}
else {
if(ma[0][0] == '1') printf("1");
for(int i = 0;i < ans.size(); i++)
printf("%d",ans[i]);
printf("\n");
}
}
return 0;
}
/*
20
2 2
11
11
3 3
001
111
101
3 3
111
111
111
3 3
000
000
000
3 3
001
011
111
*/