HEU1728 逃离迷宫

题目地址：点击打开链接

思路：用DFS容易超时，用BFS在一个方向搜到底，要多减枝

超时代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;


int visit[110][110];
char map[110][110];
int m,n,endi,endj,flag,k;
int dir[4][2] = {{-1,0,},{1,0},{0,-1},{0,1}};


void dfs(int x,int y,int direction,int sum)
{
    int i,newx,newy;
    if(flag)
        return;
    if(sum > k)
        return;
    if(x == endi && y == endj)
    {
        flag = 1;
        return;
    }
    for(i=0; i<4; i++)
    {
        newx = x + dir[i][0];
        newy = y + dir[i][1];
        if(newx > 0 && newy > 0 && newx <= m && newy <= n && map[newx][newy] != '*' && !visit[newx][newy])
        {
            visit[newx][newy] = 1;
            if(direction == i)
                dfs(newx,newy,i,sum);
            else
                dfs(newx,newy,i,sum+1);
            visit[newx][newy] = 0;
        }
    }
}
int main()
{
    int t,i,j,x1,y1;
    cin>>t;
    while(t--)
    {
        memset(visit,0,sizeof(visit));
        flag = 0;
        cin>>m>>n;
        for(i=1; i<=m; i++)
        {
            for(j=1; j<=n; j++)
            {
                cin>>map[i][j];
            }
        }
        cin>>k>>y1>>x1>>endj>>endi;
        for(i=0; i<4; i++)
        {
            dfs(x1,y1,i,0);
        }
        if(flag)
            cout<<"yes"<<endl;
        else
            cout<<"no"<<endl;
    }
    return 0;
}

看到大神用DFS秒过，减的太好了，写个思路吧，DFS不用visit数组标记也行，只要把map改一下回溯就行，减值主要靠的是把

错误代码：

#include<iostream>
#include<queue>

using namespace std;

char map[110][110];
int dir[4][2] = {{-1,0,},{1,0},{0,-1},{0,1}};
int m,n,k,endi,endj;

struct node
{
    int x;
    int y;
    int dir;
}a[10010];
bool check(int x,int y)
{
    if(x>0 && y>0 && x<=m && y<=n && map[x][y] != '*')
        return true;
    return false;
}
void bfs(int x,int y)
{
    if(x == endi && y == endj)
    {
        cout<<"yes"<<endl;
        return;
    }
    int i,newx,newy;
    queue<node> q;
    node in,out;
    in.x = x;
    in.y = y;
    in.dir = -1;
    q.push(in);
    while(!q.empty())
    {
        out = q.front();
        q.pop();
        for(i=0; i<4; i++)
        {
            newx = out.x + dir[i][0];
            newy = out.y + dir[i][1];
            while(check(newx,newy))
            {
                in.x = newx;
                in.y = newy;
                in.dir = out.dir + 1;
                q.push(in);
                map[newx][newy] = '*';//不能用这个标记已经走过，会导致，以后能走的路却不能走
                if(newx == endi && newy == endj && in.dir <= k)
                {
                    cout<<"yes"<<endl;
                    return;
                }
                newx += dir[i][0];
                newy += dir[i][1];
            }
        }
    }
    cout<<"no"<<endl;
}
int main()
{
    int t,i,j,x1,y1;
    cin>>t;
    while(t--)
    {
        cin>>m>>n;
        for(i=1; i<=m; i++)
        {
            for(j=1; j<=n; j++)
            {
                cin>>map[i][j];
            }
        }
        cin>>k>>y1>>x1>>endj>>endi;
        bfs(x1,y1);
    }
    return 0;
}

AC代码：

#include<iostream>
#include<queue>
#include<cstring>

using namespace std;

int visit[110][110];
char map[110][110];
int dir[4][2] = {{-1,0,},{1,0},{0,-1},{0,1}};
int m,n,k,endi,endj;

struct node
{
    int x;
    int y;
    int dir;
}a[10010];
bool check(int x,int y)
{
    if(x>0 && y>0 && x<=m && y<=n && map[x][y] != '*')
        return true;
    return false;
}
void bfs(int x,int y)
{
    if(x == endi && y == endj)
    {
        cout<<"yes"<<endl;
        return;
    }
    int i,newx,newy;
    queue<node> q;
    node in,out;
    in.x = x;
    in.y = y;
    in.dir = -1;
    q.push(in);
    while(!q.empty())
    {
        out = q.front();
        q.pop();
        for(i=0; i<4; i++)
        {
            newx = out.x + dir[i][0];
            newy = out.y + dir[i][1];
            while(check(newx,newy))
            {
                if(!visit[newx][newy])
                {
                    in.x = newx;
                    in.y = newy;
                    in.dir = out.dir + 1;
                    q.push(in);
                    visit[newx][newy] = 1;
                    if(newx == endi && newy == endj && in.dir <= k)
                    {
                        cout<<"yes"<<endl;
                        return;
                    }
                }
                newx += dir[i][0];
                newy += dir[i][1];
            }
        }
    }
    cout<<"no"<<endl;
}
int main()
{
    int t,i,j,x1,y1;
    cin>>t;
    while(t--)
    {
        memset(visit,0,sizeof(visit));
        cin>>m>>n;
        for(i=1; i<=m; i++)
        {
            for(j=1; j<=n; j++)
            {
                cin>>map[i][j];
            }
        }
        cin>>k>>y1>>x1>>endj>>endi;
        bfs(x1,y1);
    }
    return 0;
}

大神解题地址： 点击打开链接