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hdu--1026--bfs&&优先队列&&打印路径

时间:2014-07-24 05:01:48      阅读:362      评论:0      收藏:0      [点我收藏+]

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这题 是被我自己搞复杂了....

太SB了....

还是porker的关于输出路径的简洁 有效多了

    touch  me

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#include <iostream>
#include <cstring>
#include <queue>
#include <stack>
using namespace std;

int ans, n, m;
const int size = 110;
char maze[size][size];
bool vis[size][size];
int arr[10010];
int dir[4][2] = { 1, 0, -1, 0, 0, 1, 0, -1 };

struct data
{
    int x, y;
    int step;
    bool operator<(const data& p)const
    {
        return step>p.step;
    }
}st;

void bfs()
{
    data now, next;
    int xx, yy;
    priority_queue<data> q;
    while (!q.empty())
        q.pop();
    st.x = st.y = st.step = 0;
    q.push(st);
    vis[0][0] = true;
    while (!q.empty())
    {
        now = q.top();
        q.pop();
        if (now.x == n - 1 && now.y == m - 1)
        {
            ans = now.step;
            return;
        }
        for (int i = 0; i < 4; i++)
        {
            xx = now.x + dir[i][0];
            yy = now.y + dir[i][1];
            if (xx >= 0 && xx < n && yy >= 0 && yy < m && maze[xx][yy] != X && !vis[xx][yy])
            {
                vis[xx][yy] = true;
                next.x = xx;
                next.y = yy;
                arr[xx*m + yy] = now.x*m + now.y;
                if (maze[xx][yy] == .)
                    next.step = now.step + 1;
                else
                    next.step = now.step + (maze[xx][yy] - 0) + 1;
                q.push(next);
            }
        }
    }
    return;
}

int main()
{
    cin.sync_with_stdio(false);
    stack<int>s;
    int num1, x1, y1, num2, x2, y2;
    int k, flag;
    while (cin >> n >> m)
    {
        flag = 0;
        while (!s.empty())
            s.pop();
        memset(vis, false, sizeof(vis));
        memset(arr, -1, sizeof(arr));
        ans = 0;
        k = 1;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                cin >> maze[i][j];
            }
        }
        bfs();
        if (ans)
        {
            cout << "It takes " << ans << " seconds to reach the target position, let me show you the way." << endl;
            for (int i = n*m - 1; i != 0; i = arr[i])
            {
                s.push(i);
            }
            int prex = 0, prey = 0;
            while( !s.empty() ) 
            {
                num1 = s.top();
                s.pop();
                x1 = num1 / m;
                y1 = num1 % m;
                cout << k++ << "s:(" << prex << "," << prey << ")->(" << x1 << "," << y1 << ")" << endl;
                if (maze[x1][y1] != .) 
                {
                    for (int i = 0; i < maze[x1][y1] - 0; i++) 
                    {
                        cout << k++ << "s:FIGHT AT (" << x1 << "," << y1 << ")" << endl;
                    }
                }
                prex = x1;
                prey = y1;
            }
        }
        else
        {
            cout << "God please help our poor hero." << endl;
        }
        cout << "FINISH" << endl;
    }
    return 0;
}
View Code

这是他帮我修改后过的  贴下起初没过的。。。--纪念下自己的白痴的k ans 做法=-= 

bubuko.com,布布扣
  1 #include <iostream>
  2 #include <cstring>
  3 #include <queue>
  4 #include <stack>
  5 using namespace std;
  6 
  7 int ans , n , m;
  8 const int size = 110;
  9 char maze[size][size];
 10 bool vis[size][size];
 11 int arr[10010];
 12 int dir[4][2] = {1,0,-1,0,0,1,0,-1};
 13 
 14 struct data
 15 {
 16     int x , y;
 17     int step;
 18     bool operator<(const data& p )const
 19     {
 20         return step>p.step;
 21     }
 22 }st;
 23 
 24 void bfs( )
 25 {
 26     data now , next;
 27     int xx , yy;
 28     priority_queue<data> q;
 29     while( !q.empty() )
 30         q.pop();
 31     st.x = st.y = st.step = 0;
 32     q.push( st );
 33     vis[0][0] = true;
 34     while( !q.empty() )
 35     {
 36         now = q.top();
 37         q.pop();
 38         if( now.x == n-1 && now.y == m-1 )
 39         {
 40             ans = now.step;
 41             return;
 42         }
 43         for( int i = 0 ; i<4 ; i++ )
 44         {
 45             xx = now.x + dir[i][0];
 46             yy = now.y + dir[i][1];
 47             if( xx>=0 && xx<n && yy>=0 && yy<m && maze[xx][yy]!=X && !vis[xx][yy] )
 48             {
 49                 vis[xx][yy] = true;
 50                 next.x = xx;
 51                 next.y = yy;
 52                 arr[xx*m+yy] = now.x*m+now.y;
 53                 if( maze[xx][yy] == . )
 54                     next.step = now.step + 1;
 55                 else
 56                     next.step = now.step + (maze[xx][yy] -0)+1;
 57                 q.push(next);
 58             }
 59         }
 60     }
 61     return;
 62 }
 63 
 64 int main()
 65 {
 66     stack<int>s;
 67     int num1 , x1 , y1 , num2 , x2 , y2;
 68     int k , flag;
 69     while( cin >> n >> m )
 70     {
 71         flag = 0;
 72         while( !s.empty() )
 73             s.pop();
 74         memset( vis , false , sizeof(vis) );
 75         memset( arr , -1 , sizeof(arr) );
 76         ans = 0;
 77         k = 1;
 78         for( int i = 0 ; i<n ; i++ )
 79         {
 80             for( int j = 0 ; j<m ; j++ )
 81             {
 82                 cin >> maze[i][j];
 83             }
 84         }
 85         bfs( );
 86         if( ans )
 87         {
 88             cout << "It takes " << ans << " seconds to reach the target position, let me show you the way." << endl;
 89             for( int i = n*m-1 ; i!=-1 ; i = arr[i] )
 90             {
 91                 s.push( i );
 92             }
 93             /*
 94             while(!s.empty())
 95             {
 96                 cout << s.top() <<endl;
 97                 s.pop();
 98             }
 99             */
100             while( k<ans )
101             {
102                 num1 = s.top();
103                 x1 = num1/m;
104                 y1 = num1%m;
105                 s.pop();
106                 if( maze[x1][y1] ==. )
107                 {
108                     if( !flag )
109                     {
110                         num2 = s.top();
111                         x2 = num2/m;
112                         y2 = num2%m;
113                         cout<<k<<"s: ("<<x1<<","<<y1<<")->("<<x2<<","<<y2<<")"<<endl;
114                     }
115                     else
116                     {
117                         x2 = flag/m;
118                         y2 = flag%m;
119                         cout<<k<<"s: ("<<x2<<","<<y2<<")->("<<x1<<","<<y1<<")"<<endl;
120                         flag = 0;
121                     }
122                     k++;
123                 }
124                 else
125                 {
126                     for( int j = 0 ; j<(maze[x1][y1]-0) ; j++ )
127                     {
128                         cout<<k+j<<"s:FIGHT AT ("<<x1<<","<<y1<<")"<<endl;
129                     }
130                     k+=( maze[x1][y1]-0 );
131                     flag = num1;
132                 }
133             }
134         }
135         else
136         {
137             cout << "God please help our poor hero." << endl;
138         }
139         cout << "FINISH" << endl;
140     }
141     return 0;
142 }
View Code

洗澡去了~再看部电影 碎觉 =-=

hdu--1026--bfs&&优先队列&&打印路径,布布扣,bubuko.com

hdu--1026--bfs&&优先队列&&打印路径

标签:style   blog   http   color   os   io   

原文地址:http://www.cnblogs.com/radical/p/3864193.html

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