Hdu-5335 Walk Out (BFS+贪心)

Problem Description

In an n∗mmaze(迷失), 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(邻近的) 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(二进制的) number. Please determine theminimum(最小的) value of this number in binary system.

 

Input

The first line of the input(投入) is a singleinteger(整数)T (T=10),indicating(表明) 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(迷宫).

 

Output

For each testcase, print the answer in binary(二进制的) system. Pleaseeliminate(消除) all thepreceding(领先)0 unless the answer itself is 0 (in this case, print 0 instead).

 

Sample Input

22 211113 3001111101

 

Sample Output

111101

 

Author

XJZX

 

Source

2015 Multi-University Training Contest 4

 

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题意：给你一个n*m的零一矩阵，起点在左上角终点在右下角，没走过一步记录下脚下的数字，最后求按二进制系统 最小的路径。


分析：考虑起始位置是不是0，若是0则BFS出所有离起点曼哈顿距离最远的点否则以（1，1）作为起点，BFS最小路径。

#include <cstdio>#include <iostream>#include <cstring>using namespace std;const int f[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};char S[1005][1005];struct thing{int x,y,v,f;} q[1000007],Nq[1000007];int T,n,m,s,t,ans[2010];bool jud[1005][1005],F[2010];int main(){scanf("%d",&T);while(T--){memset(jud,0,sizeof(jud));memset(F,0,sizeof(F));scanf("%d%d",&n,&m);for(int i = 1;i <= n;i++) scanf("%s",S[i]+1);if(S[1][1] == '0'){s = 1,t = 2;q[s].x = 1,q[s].y = 1;jud[1][1] = true;int Max = 2;while(s != t){for(int k = 0;k < 4;k++){int Nx = q[s].x + f[k][0],Ny = q[s].y + f[k][1];if(Nx && Nx <= n && Ny && Ny <= m && S[Nx][Ny] == '0'){q[t].x = Nx;q[t].y = Ny;Max = max(Max,Nx+Ny);if(!jud[q[t].x][q[t].y]){jud[q[t].x][q[t].y] = true;t++;}}}s++;}t--;s = 1;bool flag = false;while(t){if(q[t].x == n && q[t].y == m) {cout<<0<<endl;flag = true;break;}if (q[t].x+q[t].y == Max){if(q[t].x < n){Nq[s].x = q[t].x+1;Nq[s].y = q[t].y;if(!jud[Nq[s].x][Nq[s].y]){Nq[s].f = 0;jud[Nq[s].x][Nq[s].y] = true;s++;}} if(q[t].y < m){Nq[s].x = q[t].x;Nq[s].y = q[t].y+1;if(!jud[Nq[s].x][Nq[s].y]){Nq[s].f = 0;jud[Nq[s].x][Nq[s].y] = true;s++; }}}t--;}if(flag) continue;t = s;s = 1;}else {s = 1,t = 2;Nq[s].x = 1;Nq[s].y = 1;Nq[s].f = 0;jud[1][1] = true;}while(s != t){if(F[Nq[s].x+Nq[s].y] && S[Nq[s].x][Nq[s].y] == '1') {s++;continue;}if(Nq[s].x < n){int Nx = Nq[s].x+1,Ny = Nq[s].y;if(!jud[Nx][Ny]){jud[Nx][Ny] = true;Nq[t].x = Nx;Nq[t].y = Ny;Nq[t].f = s;if(S[Nx][Ny] == '0') F[Nx+Ny] = true;t++;}}if(Nq[s].y < m){int Nx = Nq[s].x,Ny = Nq[s].y+1;if(!jud[Nx][Ny]){jud[Nx][Ny] = true;Nq[t].x = Nx;Nq[t].y = Ny;Nq[t].f = s;if(S[Nx][Ny] == '0') F[Nx+Ny] = true;t++;}}s++;}for(int i = t-1;i;i--) if(Nq[i].x == n && Nq[i].y == m)  { int now = i,cnt = 0;while(now){ans[++cnt] = S[Nq[now].x][Nq[now].y];now = Nq[now].f;} for(int j = cnt;j;j--) cout<<char(ans[j]);cout<<endl; break; }}}