Walk Out

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1292 Accepted Submission(s): 239

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.

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.

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

3 3

001

111

101

Sample Output

111

101

题目大意：t组数据。每组n，m分别表示行和列。下面的n行m列用0、1表示地图情况。问你从（1，1）走到（n，m）所经过的点按顺序形成的二级制数最小是多少？

解题思路：先把所有的跟入口相连的0连通块都标记上。然后找到跟这个0连通块相邻接的1.这些1是可能的出发点。再在这些1中找到离出口最近的点。从这些点只需要往右往下走。因为当你确定离出口最近的点的时候，二进制数的位数已经确定。然后再广搜，贪心0。如果广搜时发现有0，那么将所有的0入队，如果没有发现0，那么将所有的1入队。

#include<bits/stdc++.h>

using namespace std;

const int maxn=1010;

const int INF=0x3f3f3f3f;

int vis[maxn][maxn];

int Map[maxn][maxn];

int way[2*maxn];

int f[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

int n,m,k;

struct NODE{

    int x,y;

};

queue<NODE>Q;

stack<NODE>S;

int BFS(){

    NODE st,tmp;

    while(!Q.empty()){

        st=Q.front();

        Q.pop();

        int xx,yy;

        for(int i=0;i<4;i++){

            xx=st.x+f[i][0];

            yy=st.y+f[i][1];

            if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&(!vis[xx][yy])&&Map[xx][yy]==0){

                if(xx==n&&yy==m)

                    return -1;

                tmp.x=xx,tmp.y=yy;

                vis[xx][yy]=1;

                Q.push(tmp);

            }

        }

    }

}

bool check(int x,int y){

    int xx,yy;

    for(int i=0;i<4;i++){

        xx=x+f[i][0];

        yy=y+f[i][1];

        if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&vis[xx][yy]==1){

            return true;

        }

    }

    return false;

}

void Find1(int typ){

    while(!S.empty())

        S.pop();

    while(!Q.empty())

        Q.pop();

    int maxs=-INF;

    NODE st,tmp;

    if(typ==0){

        for(int i=1;i<=n;i++){

            for(int j=1;j<=m;j++){

                if(Map[i][j]==1){

                    if(check(i,j)){

                        if(i+j>maxs){

                            maxs=i+j;

                        }

                        st.x=i,st.y=j;

                        vis[i][j]=2;

                        S.push(st);

                    }

                }

            }

        }

    }else{

        st.x=1,st.y=1;

        vis[1][1]=2;

        maxs=2;

        S.push(st);

    }

    if(S.empty()==0)

        way[k++]=1;

    while(!S.empty()){

        st=S.top();

        S.pop();

        if(st.x+st.y==maxs){

            Q.push(st);

        }

    }

    while(1){

        int flag=0;

        while(!Q.empty()){

            st=Q.front();

            Q.pop();

            if(st.x==n&&st.y==m){

                return ;

            }

            int xx,yy;

            for(int i=0;i<2;i++){

                xx=st.x+f[i][0];

                yy=st.y+f[i][1];

                if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&(!vis[xx][yy])){

                    if(Map[xx][yy]==0){

                        flag=1;

                    }

                    tmp.x=xx,tmp.y=yy;

                    vis[xx][yy]=2;

                    S.push(tmp);

                }

            }

        }

        if(flag==1){

            way[k++]=0;

            while(!S.empty()){

                st=S.top();

                S.pop();

                if(Map[st.x][st.y]==0){

                    Q.push(st);

                }

            }

        }else{

            way[k++]=1;

            while(!S.empty()){

                st=S.top();

                S.pop();

                Q.push(st);

            }

        }

    }

}

int main(){

    int t;

    char str[1020];

    scanf("%d",&t);

    while(t--){

        k=0;

        memset(vis,0,sizeof(vis));

        scanf("%d %d",&n,&m);

        for(int i=1;i<=n;i++){

           scanf("%s",str+1);

            for(int j=1;j<=m;j++){

                Map[i][j]=str[j]-'0';

            }

        }

        if(n==m&&n==1&&Map[1][1]==0){

            printf("0\n");continue;

        }

        if(Map[1][1]==0){

            NODE st;

            st.x=1,st.y=1;

            vis[1][1]=1;

            while(!Q.empty())

                Q.pop();

            Q.push(st);

            if(BFS()==-1) {

                printf("0\n");

                continue;

            }

            Find1(0);

        }

        else{

            Find1(1);

        }

        for(int i=0;i<k;i++){

            printf("%d",way[i]);

        }printf("\n");

    }

    return 0;

}

/*

50

3 3

001

111

101

55

1 2

01

*/

巴特西

HDU 5335——Walk Out——————【贪心】

Walk Out

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