【每日刷题】过河卒

题目地址

https://www.luogu.com.cn/problem/P1002

解答

动态规划可以解决该问题。

为什么用动态规划？注意题目中的关键一点：只能向下或向右走。

此时可以列出表达式：record[i][j] = record[i-1][j] + record[i][j-1].

首先：不考虑“马”存在的情况，即路上没有障碍。因此可以一路走下去，直接到终点即可。

如果考虑“马”，则在“马”能设计到的范围内，将 record[i][j] 设置为 0, 这意味该点不可达。

#include 
#include 
//-----------------------
using namespace std;
//-----------------------
bool isHorseRange(const int& HorseX,const int& HorseY, const int& nowX, const int& nowY){
    if( (nowX == HorseX - 1 && nowY == HorseY - 2)
        ||  (nowX == HorseX - 2 && nowY == HorseY - 1)
        ||  (nowX == HorseX - 2 && nowY == HorseY + 1)
        ||  (nowX == HorseX - 1 && nowY == HorseY + 2)
        ||  (nowX == HorseX + 1 && nowY == HorseY + 2)
        ||  (nowX == HorseX + 1 && nowY == HorseY - 2)
        ||  (nowX == HorseX + 2 && nowY == HorseY - 1)
        ||  (nowX == HorseX + 2 && nowY == HorseY + 1)
        ||  (nowX == HorseX && nowY == HorseY)
        )
        return true;
    return false;
}
int main() {
    int Bx, By, HorseX, HorseY;
    cin>>Bx>>By>>HorseX>>HorseY;

    vector<vector<long>> record(Bx + 1, vector<long>(By + 1, 0));

    record[0][0] = !isHorseRange(HorseX, HorseY, 0, 0);
    //两条边界线上初始化
    for( int i = 1; i < Bx + 1; i++){
        if( isHorseRange(HorseX,HorseY, i, 0))
            record[i][0] = 0;
        else
            record[i][0] = record[i-1][0];
    }

    for( int j = 1; j < By + 1; j++){
        if( isHorseRange(HorseX,HorseY, 0, j))
            record[0][j] = 0;
        else
            record[0][j] = record[0][j-1];
    }
    //中间区域
    for( int i = 1; i < Bx + 1; i++){
        for( int j = 1; j < By + 1; j++){
            if( isHorseRange(HorseX, HorseY, i, j))
                record[i][j] = 0;
            else
                record[i][j] = record[i-1][j] + record[i][j-1];
        }
    }

    cout<<record[Bx][By];

return 0;}