HDU 6048 Puzzle（滑块游戏）

Puzzle

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 92    Accepted Submission(s): 48

Problem Description

A Jigsaw puzzle contains N*M-1 pieces of jigsaws in a N rows*M columns rectangular board.Each jigsaw has a distinct number from 1 to N*M-1.Li is a naughty boy,he wants to arrange the board in his unique way.At the beginning,he picks all N*M-1 jigsaws out and put them on the table and then he will put them back to the board respecting the following steps:
1.Sorting all the remaining jigsaws on the table in ascending order.
2.Picking out the 1st ,the P+1 th ,the 2*P+1 th,......the n*P+1 th jigsaws and put them back to the blank area in the board one by one from the top row to the bottom row,from the left column to the right column.
3.if there are jigsaws remained on the table,back to step 1.
After he arranging the board,it’s obvious that there’s only one blank area located at the bottom-right corner.
Your task is to make the numbers on jigsaws sorted with every row and every column in ascending order(From left to right,top to bottom),and the blank area should be located at the bottom-right corner in the end.Each step you can move the blank area’s neighboring jigsaws(which share a common side with the blank area) towards the blank area.It’s really a difficult question,so you need to write a program to judge whether it is possible to complete the task.

 

Input

The first line contains an integer T(T<=100),which represents the number of test cases.
Following T lines,each line contains three integers N,M,P(2<=N,M<=1000;1<=P<=N*M-2).

 

Output

For each test case,print “YES” in a separate line if it is possible to complete the task ,otherwise please print “NO”.

 

Sample Input

33 2 33 2 4999 999 1

 

Sample Output

YESNOYES

 

Source

2017 Multi-University Training Contest - Team 2 

 

Recommend

liuyiding

 

题目大意：

     给你一个N*M的矩阵，从上到下从左到右，依次填1, 2, 3,...右下角为空，每次可以把一个和空格有公共边的块和空格交换。按照特定方式打乱，问是否有解。

解题思路：

    纯考察思维和找规律的一道题，最终结论就是按照第一行，第二行，第三行，。。。合并成一个一维数组，数组的逆序对数为偶数则有解，否则无解。证明比较长，官方题解写的很详细，这里就引用官方题解：

AC代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <ctime>#include <vector>#include <queue>#include <stack>#include <deque>#include <string>#include <map>#include <set>#include <list>using namespace std;#define INF 0x3f3f3f3f#define LL long long#define fi first#define se second#define mem(a,b) memset((a),(b),sizeof(a))int N,M,P;int main(){    int T_T;    scanf("%d",&T_T);    while(T_T--)    {        scanf("%d%d%d",&N,&M,&P);        int num=N*M-1, cnt=0;//剩下的数字数，逆序对数        while(num>P)        {            int n=(num-1)/P+1;//这一轮拿出的数字            cnt+=n*(n-1)/2*(P-1);//逆序对数            num-=n;        }        puts(cnt&1?"NO":"YES");    }        return 0;}