【hdu 1112】The Proper Key

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 487 Accepted Submission(s): 157

Problem Description

Many people think that Tetris was invented by two Russian programmers. But that is not the whole truth. The idea of the game is very old – even the Egyptians had something similar. But they did not use it as a game. Instead, it was used as a very complicated lock. The lock was made of wood and consisted of a large number of square fields, laid out in regular rows and columns. Each field was either completely filled with wood, or empty. The key for this lock was two-dimensional and it was made by joining square parts of the same size as the fields of the lock. So they had a 2D lock and 2D key that could be inserted into the lock from the top. The key was designed so that it was not possible to move it upwards. It could only fall down and it could slide sideways – exactly like in a Tetris game. The only difference is that the key could not be rotated. Rotation in Tetris is really a Russian invention.

The entry gate into the Pyramid has such a lock. The ACM archaeologists have found several keys and one of them belongs to the lock with a very high probability. Now they need to try them out and find which one to use. Because it is too time-consuming to try all of them, it is better to begin with those keys that may be inserted deeper into the lock. Your program should be able to determine how deep a given key can be inserted into a given lock.

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers R and C (1 <= R,C <= 100) indicating the key size. Then exactly R rows follow, each containing C characters. Each character is either a hash mark (#) or a period (.). A hash mark represents one square field made of wood; a period is an empty field. The wooden fields are always connected, i.e. the whole key is made of one piece. Moreover, the key remains connected even if we cut off arbitrary number of rows from its top. There is always at least one non-empty field in the top-most and bottom-most rows and the left-most and right-most columns.

After the key description, there is a line containing two integers D and W (1 <= D <= 10000, 1 <= W <= 1000). The number W is the lock width, and D is its depth. The next D lines contain W characters each. The character may be either a hash mark (representing the wood) or a period (the free space).

Output

Your program should print one line of output for each test case. The line should contain the statement “The key falls to depth X.”. Replace X with the maximum depth to which the key can be inserted by moving it down and sliding it to the left or right only. The depth is measured as the distance between the bottom side of the key and the top side of the lock. If it is possible to move the key through the whole lock and take it away at the bottom side, output the sentence “The key can fall through.”.

Sample Input
4
2 4
#.##
###.
3 6
#....#
#....#
#..###
2 3
##.
.##
2 7
#.#.#.#
.#.#.#.
1 1
#
1 10
###....###
3 2
##
.#
.#
1 5
#.#.#

Sample Output
The key falls to depth 2.
The key falls to depth 0.
The key can fall through.
The key falls to depth 2.

【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=1112

【题解】



让你那上面那个钥匙去插入下面那个锁里面;

问你这个钥匙能插多深.

这里的深度是指钥匙最后的底部和锁的上部的高度差;

我们首先可以让锁和钥匙左对齐;

即它们两个都靠左摆好;

设钥匙的高为n宽为m;

设锁的高为nn宽为mm

则钥匙能够往右移动的最大限度是mm-m;

(这道题的设置是,如果钥匙的宽比锁长就直接输出0);



也就是说你一开始有mm-m+1个位置可以插入;

而且钥匙插入锁里面之后可以左右移动(也有mm-m+1个位置可供移动);

深度的最大值为n+nn-1,一旦大于这个值就完全插入了;

代码的模拟过程是,一个单位一个单位地把钥匙往下移动;

先看看锁能在第一层的那mm-m+1个口中的哪一些口插入;

用can[r][mm-m]表示;->bool数组;

然后从第二层开始,就可能有横移操作了;

即可能can[1][j]这个状态(就是说从第j个口不能插入);

但是can[2][j-1]这个状态(第j-1个口,能一直插到第2行)可行;

那么就能从can[2][j-1]这个状态表示的情况,再右移一位;那么就能达到

can[2][j]了;这时can[2][j]不是表示从第j个口直接往下移动2层;

而是先从第j-1个口往下移动两层,再往右移动一层;

(我说的口是下面这个图所示的意思,一共mm-m+1个口)



然后如果新获得了状态can[i][j]

那么如果can[i][j-1]为false;

则让j减去1(在循环体里面要减去2,因为有个j++还没执行);

然后再看看can[i][j-1]能不能因为can[i][j]变成true了,然后can[i][j-1]本身也变成true;

(即左移)

这里判断能不能到can[i][j]这个状态.

可以看一下是钥匙的一部分的地方(x,y)

其(x+i,y+j)是不是锁;

如果是钥匙的地方(x,y),对应的(x+i,y+j)都为空,就表示这个状态可以装得下这把钥匙;

同时因为can[i-1][j]或can[i][j-1]或can[i][j+1]为true;

就说明这个位置能通过移动和外界联系;则可行;

具体的看代码吧;



【完整代码】

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 100+100;

int T;
int n,m,nn,mm;
bool key[MAXN][MAXN],loc[10010][1010],can[10010][1010];
char s[1000+100];

bool in(int px,int py)
{
    for (int i = 1;i <= n;i++)
        for (int j = 1;j <= m;j++)
            if (key[i][j] && loc[i+px][j+py])
                return false;
    return true;
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    cin >> T;
    while (T--)
    {
        for (int i = 1;i <= 100;i++)
            for (int j = 1;j <= 100;j++)
                key[i][j] = false;
        for (int i = 1;i <= 10000;i++)
            for (int j = 0;j <= 1000;j++)
                loc[i][j] = can[i][j] = false;
        cin >> n >> m;
        for (int i = 1;i <= n;i++)
        {
            scanf("%s",s+1);
            for (int j = 1;j <= m;j++)
                if (s[j]=='#')
                    key[i][j] = true;
        }
        cin >> nn >> mm;
        nn+=n;
        for (int i = n+1;i <= nn;i++)
        {
            scanf("%s",s+1);
            for (int j = 1;j <= mm;j++)
                if (s[j]=='#')
                    loc[i][j] = true;
        }
        for (int i = 0;i <= mm;i++)
            can[0][i] = true;
        int i;
        for (i = 1;i <= nn;i++)
        {
            bool flag = false;
            for (int j = 0;j <= mm-m;j++)
            {
                if (can[i][j]) continue;
                bool judge1 = (j-1>=0 && can[i][j-1]);
                bool judge2 = (j+1<=mm-m && can[i][j+1]);
                bool judge3 = can[i-1][j];
                if (judge1 || judge2 || judge3)
                {
                    if (in(i,j))
                    {
                       flag = true;
                       can[i][j] = true;
                       if (j-1>=0 && !can[i][j-1])
                            j-=2;
                    }
                }
            }
            if (!flag)
                break;
        }
        if (i>nn)
            puts("The key can fall through.");
        else
            printf("The key falls to depth %d.\n",i-1);
    }
    return 0;
}