SGU 139. Help Needed! 逆序数,奇偶性,分析 难度:0

139. Help Needed!

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB

 

Little Johnny likes puzzles a lot! Just a few days ago, he found out about the 'traditional' 4x4 puzzle. For this puzzle, you have all the numbers from 0 to 15 arranged in 4 rows and 4 columns. You are allowed to switch two adjacent elements (horizontally or vertically), only if one of them has the value 0. The purpose of the puzzle is to reach the following final state:

                             1  2  3  4 
                             5  6  7  8 
                             9 10 11 12 
                            13 14 15  0

Given the initial state of the puzzle, you have to decide whether there exists a sequence of moves which brings the puzzle into the final state.

 

Input

The input will consist of  4 lines, each of them containing 4 integers, describing the initial state of the puzzle.

 

Output

For every initial state, you should print "YES" if the final state can be reached after several moves or "NO", if such a thing is impossible.

 

Sample Input #1

1 2 3 4
5 6 7 8
9 10 11 0
13 14 15 12

Sample Output #1

YES

Sample Input #2

2 1 3 4
5 6 7 8
9 10 11 12
0 13 14 15

Sample Output #2

NO
首先得到逆序数状态,移动0可以+1,-1逆序对或者+3,-3逆序对.只要逆序数奇偶性和必须走的步数相同,那么就一定能走到逆序对为0的情况(设0为16)

#include <cstdio>
using namespace std;
int des[4][4];
int main(){
    int aim=0;
    for(int i=0;i<4;i++){
        for(int j=0;j<4;j++){
            scanf("%d",des[i]+j);
            if(des[i][j]==0){
                aim=6-i-j;
                des[i][j]=16;
            }
        }
    }
    for(int i=0;i<16;i++){
        for(int j=0;j<i;j++){
            if(*((int *)des+i)<*((int *)des+j)){
                aim++;
            }
        }
    }
    if(aim&1)puts("NO");
    else puts("YES");
    return 0;
}