ZOJ-1002-Fire Net

这道题把city类的成员函数定义的格式写得特复杂，其实当时是想分出头文件和source文件才写得这样麻烦的，后来为了提交OJ，也只好这样了，所以蛮难看的。。。不过尽管代码难看，函数的功能还是比较清晰，解释也在注释里了，就不多说了，写这个程序的时候确实也好久都没碰C++了，所以凑合看吧。

总得来说，这道题的好玩大于难度。

Fire NetTime limit: 1 Seconds   Memory limit: 32768K  
Total Submit: 5221   Accepted Submit: 1702  

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 is legal 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 integer n 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

Problem Source: Zhejiang University Local Contest 2001

#include <iostream>using namespace std;static int num,maxbhnum=0,nowmax;class city{public:            int FlagNum;//方阵的行（列）数    int Wall;//黑方块（墙）的个数    char OrPattern[5][5];//以数字表示的给定布局    int ReadPattern();//1.从文件读取原始布局到OrPattern,                      //2.并转换出数字表示的布局Pattern,                      //3.读出方阵的行（列）数FlagNum,              //4.返回墙Wall的数目    int Result();//递归函数，生成各种blockhouse的布局状态，以全局变量num监测放置blockhouse的数目    bool can();//在Result中判断一个临时生成的blockhouse布局矩阵是否合法};int city::ReadPattern(){            int count=0,i,k;    cin>>city::FlagNum;    if(0!=city::FlagNum)    {        for(i=0;i<city::FlagNum;i++)        {                cin>>city::OrPattern[i];            for(k=0;k<city::FlagNum;k++)            {                if('.'==city::OrPattern[i][k])                {                    city::OrPattern[i][k]=0;                }                else if('X'==city::OrPattern[i][k])                {                    city::OrPattern[i][k]=1;                    count++;                }                else                {                    return 27;                    break;                }            }        }    }    else    {        return -1;    }    return count;}int city::Result(){    int i,k;            for(i=0;i<city::FlagNum;i++)    {        for(k=0;k<city::FlagNum;k++)        {            if(0==OrPattern[i][k])            {                OrPattern[i][k]=2;num++;                if(true==city::can() && num<=maxbhnum)                {                    if(nowmax<num)                        nowmax=num;                    city::Result();                }                OrPattern[i][k]=0;num--;            }        }            }        return nowmax;    }bool city::can(){    int i,k,rflag,cflag;    int Flag=1;    for(i=0;i<city::FlagNum;i++)    {        rflag=1;        for(k=0;k<city::FlagNum;k++)        {                        if(2==OrPattern[i][k])            {                if(rflag==1)                {                    rflag=0;                }                else                {                    rflag=-1;                    Flag=-1;                }            }            else if(1==OrPattern[i][k])            {                rflag=1;            }            else if(0==OrPattern[i][k])            {            }        }    }        for(k=0;k<city::FlagNum;k++)    {        cflag=1;        for(i=0;i<city::FlagNum;i++)        {                        if(2==OrPattern[i][k])            {                if(cflag==1)                {                    cflag=0;                }                else                {                    cflag=-1;                    Flag=-1;                }            }            else if(1==OrPattern[i][k])            {                cflag=1;            }            else if(0==OrPattern[i][k])            {            }        }    }        if((rflag>=0) && (cflag>=0) && Flag>=0)        return true;    else        return false;}//rflag行标志,cflag列标志//1表示此行（列）已设置blockhouses//0表示此行（列）至结尾处无blockhouses，或在结尾前被墙阻挡//-1表示此行（列）的blockhouses设置不合法，失败。int main(){    city cit;    cit.Wall=cit.ReadPattern();        while(cit.Wall!=-1 && cit.Wall!=27)    {        num=0;        nowmax=0;        maxbhnum=cit.FlagNum*cit.FlagNum-cit.Wall;        cit.Result();        cout<<nowmax<<endl;        cit.Wall=cit.ReadPattern();    }}