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            posts - 3,  comments - 1,  trackbacks - 0
            這題沒(méi)把我弄瘋了.一個(gè)小時(shí)寫(xiě)完,改了2個(gè)小時(shí)...題目給的數(shù)據(jù)太弱了,需要自己寫(xiě)一些數(shù)據(jù)來(lái)驗(yàn)證...在這里給大家提供些數(shù)據(jù)

            題目
            Maze
            Time Limit: 2000MS Memory Limit: 65536K
            Total Submissions: 1205 Accepted: 399

            Description

            Acm, a treasure-explorer, is exploring again. This time he is in a special maze, in which there are some doors (at most 5 doors, represented by 'A', 'B', 'C', 'D', 'E' respectively). In order to find the treasure, Acm may need to open doors. However, to open a door he needs to find all the door's keys (at least one) in the maze first. For example, if there are 3 keys of Door A, to open the door he should find all the 3 keys first (that's three 'a's which denote the keys of 'A' in the maze). Now make a program to tell Acm whether he can find the treasure or not. Notice that Acm can only go up, down, left and right in the maze.

            Input

            The input consists of multiple test cases. The first line of each test case contains two integers M and N (1 < N, M < 20), which denote the size of the maze. The next M lines give the maze layout, with each line containing N characters. A character is one of the following: 'X' (a block of wall, which the explorer cannot enter), '.' (an empty block), 'S' (the start point of Acm), 'G' (the position of treasure), 'A', 'B', 'C', 'D', 'E' (the doors), 'a', 'b', 'c', 'd', 'e' (the keys of the doors). The input is terminated with two 0's. This test case should not be processed.

            Output

            For each test case, in one line output "YES" if Acm can find the treasure, or "NO" otherwise.

            Sample Input

            4 4
            S.X.
            a.X.
            ..XG
            ....
            3 4
            S.Xa
            .aXB
            b.AG
            0 0
            

            Sample Output

            YES
            NO
             
            數(shù)據(jù):
            5 5
            S....
            XXAXa
            GX..X
            .X...
            .....

            20 20
            S..................a
            aXXXXXXXXXXAXXXXXXX.
            .X........bb......X.
            .XbXXXXXXXXXXXXXX.X.
            .X.X.....c......X.X.
            aXbX.XXXXXXXXXX.X.X.
            .X.X.X........X.X.X.
            .X.X.X.XDXXXX.X.X.X.
            .X.X.X.X..XXX.X.X.X.
            .X.X.X.X.XG.X.X.X.X.
            .X.XcX.X.XXEX.CeX.X.
            .X.X.X.X.e..X.X.X.X.
            .X.X.X.XXXXXX.X.X.X.
            .X.X.X........X.X.X.
            .X.X.XXXXXXXXXX.X.X.
            .X.X..c.........X.X.
            .X.XXXXXXXBXXXXXX.X.
            .X........b.......X.
            .XXXXXXXXXXXXXXXXXX.
            .d..e...a........a..
            主要思想,先找鑰匙..搜索一遍,得到能找到的鑰匙,然后開(kāi)門(mén).把能開(kāi)的門(mén)都打開(kāi)..打開(kāi)門(mén)之后再找鑰匙,然后在開(kāi)門(mén).
            直到找到G..
             
            代碼如下:

            Source Code

            Problem: 2157 User: luoguangyao
            Memory: 276K Time: 0MS
            Language: C++ Result: Accepted
            • Source Code
                1#include <iostream>
                2
                3using namespace std;
                4
                5int m;
                6int n;
                7char map[22][22];
                8int allkey[6= {0};
                9int key[6= {0};
               10int mark[22][22= {0};
               11int mark2[22][22= {0};
               12int markkey[22][22= {0};
               13int lock = 1;
               14int kk = 0;
               15
               16void FindKey(int x,int y)
               17{
               18    if ((map[x][y] >= 'a'&& map[x][y] <= 'e')
               19            && markkey[x][y] != 1)
               20    {
               21        key[map[x][y] - 'a']++;
               22
               23        markkey[x][y] = 1;
               24    }

               25
               26    mark2[x][y] = 1;
               27
               28    if (mark2[x + 1][y] != 1 && x + 1 < m && map[x + 1][y] != 'X' && map[x + 1][y] != 'A' && map[x + 1][y] != 'B' 
               29        && map[x + 1][y] != 'C' && map[x + 1][y] != 'D' && map[x + 1][y] != 'E')
               30    {
               31        FindKey(x + 1 , y);
               32    }

               33
               34    if (mark2[x - 1][y] != 1 && x - 1 >= 0 && map[x - 1][y] != 'X' && map[x - 1][y] != 'A' && map[x - 1][y] != 'B'
               35        && map[x - 1][y] != 'C' && map[x - 1][y] != 'D' && map[x - 1][y] != 'E')
               36    {
               37        FindKey(x - 1 , y);
               38    }

               39
               40    if (mark2[x][y + 1!= 1 && y + 1 < n && map[x][y + 1!= 'X'  && map[x][y + 1!= 'A' && map[x][y + 1!= 'B'
               41        && map[x][y + 1!= 'C' && map[x][y + 1!= 'D' && map[x][y + 1!= 'E' )
               42    {
               43        FindKey(x , y + 1);
               44    }

               45
               46    if (mark2[x][y - 1!= 1 && y - 1 >= 0 && map[x][y - 1!= 'X'  && map[x][y - 1!= 'A' && map[x][y - 1!= 'B' 
               47        && map[x][y - 1!= 'C' && map[x][y - 1!= 'D' && map[x][y - 1!= 'E')
               48    {
               49        FindKey(x , y - 1);
               50    }

               51}

               52
               53void Findroute(int x,int y)
               54{
               55    if (map[x][y] == 'G')
               56    {
               57        lock = 0;
               58    }

               59
               60    FindKey(x , y);
               61
               62    mark[x][y] = 1;
               63
               64    if (mark[x + 1][y] != 1 && x + 1 < m && map[x + 1][y] != 'X')
               65    {
               66        if (map[x + 1][y] >= 'A' && map[x + 1][y] <= 'E')
               67        {
               68            if (key[map[x + 1][y] - 'A'!= 0 && allkey[map[x + 1][y] - 'A'== key[map[x + 1][y] - 'A'])
               69            {
               70                Findroute(x + 1 , y);
               71            }

               72        }

               73        else
               74        {
               75            Findroute(x + 1 , y);
               76        }

               77    }

               78
               79    if (mark[x - 1][y] != 1 && x - 1 >= 0 && map[x - 1][y] != 'X')
               80    {
               81        if (map[x - 1][y] >= 'A' && map[x - 1][y] <= 'E')
               82        {
               83            if (key[map[x - 1][y] - 'A'!= 0 && key[map[x - 1][y] - 'A'== allkey[map[x - 1][y] - 'A'])
               84            {
               85                Findroute(x - 1 , y);
               86            }

               87        }

               88        else
               89        {
               90            Findroute(x - 1 , y);
               91        }

               92    }

               93
               94    if (mark[x][y - 1!= 1 && y - 1 >= 0 && map[x][y - 1!= 'X')
               95    {
               96        if (map[x][y - 1>= 'A' && map[x][y - 1<= 'E')
               97        {
               98            if (key[map[x][y - 1- 'A'!= 0 && allkey[map[x][y - 1- 'A'== key[map[x][y - 1- 'A'])
               99            {
              100                Findroute(x , y - 1);
              101            }

              102        }

              103        else
              104        {
              105            Findroute(x , y - 1);
              106        }

              107    }

              108
              109    if (mark[x][y + 1!= 1 && y + 1 < n && map[x][y + 1!= 'X')
              110    {
              111        if (map[x][y + 1>= 'A' && map[x][y + 1<= 'E')
              112        {
              113            if (key[map[x][y + 1- 'A'!= 0 && allkey[map[x][y + 1- 'A'== key[map[x][y + 1- 'A'])
              114            {
              115                Findroute(x , y + 1);
              116            }

              117        }

              118        else
              119        {
              120            Findroute(x , y + 1);
              121        }

              122    }

              123
              124}

              125
              126int main()
              127{
              128    int i;
              129    int j;
              130
              131    while (cin >> m >> n)
              132    {
              133        int px = -1;
              134        int py = -1;
              135        int gx = -1;
              136        int gy = -1;
              137
              138        memset(key,0,sizeof(key));
              139        memset(allkey,0,sizeof(allkey));
              140        memset(map,'\0',sizeof(map));
              141        memset(mark,0,sizeof(mark));
              142        memset(mark2,0,sizeof(mark2));
              143        memset(markkey,0,sizeof(markkey));
              144        lock = 1;
              145
              146        if (m == 0 && n == 0)
              147        {
              148            break;
              149        }

              150
              151        for (i = 0; i < m ; ++i)
              152        {
              153            for (j = 0; j < n; ++j)
              154            {
              155                cin >> map[i][j];
              156
              157                if (map[i][j] >= 'a' && map[i][j] <= 'e')
              158                {
              159                    allkey[map[i][j] - 'a']++;
              160                }

              161
              162                if (map[i][j] == 'S')
              163                {
              164                    px = i;
              165                    py = j;
              166                }

              167
              168                if (map[i][j] == 'G')
              169                {
              170                    gx = i;
              171                    gy = j;
              172                }

              173            }

              174        }

              175
              176        if (px == -1 || py == -1 || gx == -1 || gy == -1)
              177        {
              178            cout << "NO" << endl;
              179            continue;
              180        }

              181
              182        Findroute(px,py);
              183
              184        if (lock == 1)
              185        {
              186            cout << "NO" << endl;
              187        }

              188        else
              189        {
              190            cout << "YES" << endl;
              191        }

              192    }

              193
              194    return 0;
              195}

              196
            posted on 2009-03-07 15:14 生活要低調(diào) 閱讀(1245) 評(píng)論(0)  編輯 收藏 引用

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