浙大 ZOJ 1002 Fire Net

Fire Net

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Time Limit: 2 Seconds                                    Memory Limit: 65536 KB                            

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Suppose that we have a square city with straight streets.  A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall.

A blockhouse is a small castle that has four openings through which  to shoot.  The four openings are facing North, East, South, and West, respectively.  There will be one machine gun shooting through each opening. 

Here we assume that a bullet is so powerful that it can run across  any distance and destroy a blockhouse on its way.  On the other hand, a wall is so strongly built that can stop the bullets.

The goal is to place as many blockhouses in a city as possible so that no two can destroy each other.  A configuration of blockhouses islegal provided that no two blockhouses are on the same horizontal row or vertical column in a map unless there is at least one wall separating them.  In this problem we will consider small square  cities (at most 4x4) that contain walls through which bullets cannot run through.  

The following image shows five pictures of the same board.  The first picture is the empty board, the second and third pictures show legal configurations, and the fourth and fifth pictures show illegal configurations. For this board, the maximum number of blockhouses in a legal configuration is 5; the second picture shows one way to do it, but there are several other ways.

Your task is to write a program that, given a description of a map,  calculates the maximum number of blockhouses that can be placed in the city in a legal configuration.

The input file contains one or more map descriptions, followed by a line containing the number 0 that signals the end of the file.  Each map description begins with a line containing a positive integern that is the size of the city; n will be at most 4. The next n lines each describe one row of the map, with a '.' indicating an open space and an uppercase 'X' indicating a wall.  There are no spaces in the input file.

For each test case, output one line containing the maximum number of blockhouses that can be placed in the city in a legal configuration.

Sample input:

4.X......XX......2XX.X3.X.X.X.X.3....XX.XX4................0

Sample output:

51524

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这个题让我更加理解了深搜，他会检索每一条可能存在的路，并且通过输出中间变化矩阵 深刻理解了回溯原理。太他妈巧妙了

#include<iostream>using namespace std;char cnt[5][5];int N,M,dic;// 设为全局量着实非常方便bool placed(int x,int y){    if(cnt[x][y]=='X'||cnt[x][y]=='Z')//若过这个结点本身就是不是放置碉堡的，那么直接中断此路      return 0;    for(int i=x; i<N; i++) //这样检查x方向是否满足放置条件，检查方法很巧妙，如果在没有遇到wall之前，发现又出现碉堡，否掉，如果    {                      //遇到墙，说明在这之前没有遇到碉堡，那么说明x方向右半部分已经满足条件，直接break        if(cnt[x][i]=='Z')            return 0;        if(cnt[x][i]=='X')            break;    }    for(int q=x; q>=0; q--)//接下来同理咯~    {        if(cnt[x][q]=='Z')            return 0;        if(cnt[x][q]=='X')            break;    }    for(int j=y; j<N; j++)    {        if(cnt[j][y]=='Z')            return 0;        if(cnt[j][y]=='X')            break;    }    for(int P=y; P>=0; P--)    {        if(cnt[P][y]=='Z')            return 0;        if(cnt[P][y]=='X')            break;    }    return 1; //如果没有在两个方向上检查到，碉堡或着wall那么更加完全能够放置的啦。}void dfs()    //深搜{    if(M<dic)        M=dic;    for(int i=0; i<N; i++)        for(int j=0; j<N; j++)        {            if(placed(i,j))            {                dic++;                cnt[i][j]='Z';                dfs();                cnt[i][j]='.';                dic--;            }        }}int main(){    while(cin>>N,N)    {        dic=M=0;        for(int i=0; i<N; i++)            for(int j=0; j<N; j++)            {                cin>>cnt[i][j];            }        dfs();        cout<<M<<endl;    }    return 0;}/**2.XX.*/