ZOJ-1002 回溯

1. 题目描述： 类似于八皇后问题，不同之处是斜线可以放，同时加了wall，以阻止bullet的穿透

 

2. 算法思想：回溯法，从第一位置开始，判断其是否满足约束条件，如果满足则放置blockhouse；如果不满足，则回溯下一个位置。

 

3. 算法代码：

 

3.1 举例：

    以4*4地图为例,放置有3个Wall,

 

  

 

     

 3.2 算法代码

#include <iostream>
using namespace std;
const int N=4;  //map size
char m[N][N]; //map definition
int totalNum=0;  //the max blockhouses can be put
bool IsOk(int r,int c)  //check whether the blockhouse can be put or not
{
    int i;
    
    //check whether bullet has been set at the same row or not
    for(i=c-1;i>=0;--i)
    {
        if(m[r][i] == 'O') //blockhouse has been set at the same row
            return false;
        if(m[r][i] == 'X')  //wall has been set,so blockhouse can be put
            break;
    }
    //check whether bullet has been set at the same col or not
    for(i=r-1;i>=0;--i)
    {
        if(m[i][c] == 'O') //blockhouse has been set at the same col
            return false;
        if(m[i][c] == 'X') //wall has been set,so blockhouse can be put
            break;
    }
    return true;
}
//back track for all the city,find out all possible number of blockhouses which can be set
void  BackTrack(int k, int cnt,int n)  
{
    int row,col;
    if(k==n*n )
    {
        if(cnt>totalNum)  //arrive the end
        {
            totalNum=cnt;  //get the final number of blockhouses
            return;
        }        
    }
    else
    {
        row = k/n; //row number
        col = k%n; //col number
        if(m[row][col]=='.' && IsOk(row,col)) //blockhouse can be set
        {
            m[row][col] = 'O'; //change its status
            BackTrack(k+1,cnt+1,n);    //backtrack next block,
            m[row][col] = '.'; 
        }
        BackTrack(k+1,cnt,n); 
    }
}

int main()
{
    int dn;
    while(cin>>dn && dn)
    {
        int i,j;
        for(i=0;i!=dn;++i)
            for(j=0;j!=dn;++j)
                cin>>m[i][j];
        totalNum=0;
        BackTrack(0,0,dn);
        cout<<totalNum<<endl;
    }
    return 0;
}