【题解】P2324 [SCOI2005]骑士精神

·有关IDA*

是带有估值函数的迭代加深搜索，表现出出色的效率。

估值函数可以简单的定义为「已经到位的骑士的个数」。

然后就是普通的迭代加深了。

——Loli

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <fstream>
#define re register
#define GC getchar()
#define Clean(X,K) memset(X,K,sizeof(X))
#define Jud(X,Y) (X<0||X>=N||Y<0||Y>=N)
using namespace std ;
int Qread () {
    int X = 0 ;
    char C = GC ;
    while (C > ‘9‘ || C < ‘0‘) C = GC ;
    while (C >=‘0‘ && C <=‘9‘) {
        X = X * 10 + C - ‘0‘ ;
        C = GC ;
    }
    return X ;
}
const int Dx[8] = {1,1,-1,-1,2,2,-2,-2},Dy[8] = {2,-2,2,-2,1,-1,1,-1},N = 5 , Aim[N + 1][N + 1] = {
    {1,1,1,1,1,-1},
    {0,1,1,1,1,-1},
    {0,0,2,1,1,-1},
    {0,0,0,0,1,-1},
    {0,0,0,0,0,-1},
    {-1,-1,-1,-1,-1,-1}
};
int A[N + 1][N + 1] , Sx , Sy ;
int G () {
    int Ans = 0 ;
    for (re int i = 0 ; i < N; ++ i) for (re int j = 0 ; j < N; ++ j) if (A[i][j] != Aim[i][j]) ++ Ans ;
    return Ans ;
}
int Ans , Fl = 0 ;
void DFS (int X , int Y , int Stp) {
    int Star = G() ;
    if (Stp + Star > Ans + 1) return ;
    if (Fl) return ;
    if (Star == 0) {
        Fl = 1 ;
        return ;
    }
    for (re int i = 0 ; i < 8 ;++ i) {
        int Tx = Dx[i] + X , Ty = Dy[i] + Y ;
        if (Jud(Tx , Ty)) continue ;
        swap(A[X][Y] , A[Tx][Ty]) ;
        DFS (Tx , Ty , Stp + 1) ;
        swap (A[X][Y] , A[Tx][Ty]) ;
    }
}
int main () {
//    freopen ("P2324.in" , "r" , stdin) ;
    int Times = Qread () ;
    while (Times -- ) {
        for (re int i = 0 ; i < N; ++ i) {
            for (re int j = 0 ; j < N; ++ j) {
                char C = GC ;
                while (C != ‘*‘ && C != ‘0‘ && C != ‘1‘) C = GC ;
                if (C == ‘1‘) A[i][j] = 1 ;
                if (C == ‘0‘) A[i][j] = 0 ;
                if (C == ‘*‘) {
                    A[i][j] = 2 ;
                    Sx = i , Sy = j ;
                }
            }
        }
        Fl = 0 ;
        for (Ans = 0 ; Ans <= 15 ; ++ Ans) {
            DFS (Sx , Sy , 0) ;
            if (Fl)  {
                cout << Ans <<endl;
                break ;
            }
        }
        if (!Fl)cout << -1 <<endl;
    }
    fclose (stdin) , fclose (stdout) ;
    return 0 ;
}

Thanks！

原文地址：https://www.cnblogs.com/bj2002/p/10573908.html