hdu 4772 Zhuge Liang's Password【模拟】水题

Zhuge Liang's Password

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2074    Accepted Submission(s): 1365

Problem Description

In the ancient three kingdom period, Zhuge Liang was the most famous and smart military leader. His enemy was Sima Yi, the military leader of Kingdom Wei. Sima Yi always looked stupid when fighting against Zhuge Liang. But it was Sima Yi who laughed to the end. 
Zhuge Liang had led his army across the mountain Qi to attack Kingdom Wei for six times, which all failed. Because of the long journey, the food supply was a big problem. Zhuge Liang invented a kind of bull-like or horse-like robot called "Wooden Bull & Floating Horse"(in abbreviation, WBFH) to carry food for the army. Every WBFH had a password lock. A WBFH would move if and only if the soldier entered the password. Zhuge Liang was always worrying about everything and always did trivial things by himself. Since Ma Su lost Jieting and was killed by him, he didn't trust anyone's IQ any more. He thought the soldiers might forget the password of WBFHs. So he made two password cards for each WBFH. If the soldier operating a WBFH forgot the password or got killed, the password still could be regained by those two password cards.
Once, Sima Yi defeated Zhuge Liang again, and got many WBFHs in the battle field. But he didn't know the passwords. Ma Su's son betrayed Zhuge Liang and came to Sima Yi. He told Sima Yi the way to figure out the password by two cards.He said to Sima Yi: 
"A password card is a square grid consisting of N×N cells.In each cell,there is a number. Two password cards are of the same size. If you overlap them, you get two numbers in each cell. Those two numbers in a cell may be the same or not the same. You can turn a card by 0 degree, 90 degrees, 180 degrees, or 270 degrees, and then overlap it on another. But flipping is not allowed. The maximum amount of cells which contains two equal numbers after overlapping, is the password. Please note that the two cards must be totally overlapped. You can't only overlap a part of them."
Now you should find a way to figure out the password for each WBFH as quickly as possible.

Input

There are several test cases.
In each test case:
The first line contains a integer N, meaning that the password card is a N×N grid(0<N<=30).
Then a N×N matrix follows ,describing a password card. Each element is an integer in a cell. 
Then another N×N matrix follows, describing another password card. 
Those integers are all no less than 0 and less than 300.
The input ends with N = 0

Output

For each test case, print the password.

Sample Input

2

1 2

3 4

5 6

7 8

2

10 20

30 13

90 10

13 21

0

Sample Output

0

2

Source

2013 Asia Hangzhou Regional Contest

 

题目大意：

给你两个N*N的矩阵，第一个矩阵可以不旋转，可以旋转90度，可以旋转180度，也可以旋转270度，问旋转完之后，两个矩阵最多能够有多少个位子的值相等。

思路：

暴力模拟即可。其实旋转第一个矩阵，并不一定需要将第一个矩阵真的旋转处理，其实直接将其翻转来看即可。

比如：

j 1 23 i    1a11a12a13 2a21a22a23 3a31a32a33上图翻译成代码之后：

        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                printf("%d ",a[i][j]);            }            printf("\n");        }

逆序旋转90度之后得到：

i 3 21 j    1a31a21a11 2a32a22a12 3a33a23a13

那么将其输出顺序按照此图修改即可。

上图翻译成代码之后：

        for(int j=0;j<n;j++)//x,y初始化为0        {            for(int i=n-1;i>=0;i--)            {                printf("%d ",a[x][y]);                y++;                if(y==n)                {                    y=0;                    x++;                }            }            printf("\n");        }

那么按照这个思路，将图旋转三次，每一次都判断一下即可：

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int a[45][45];int b[45][45];int temp[45][45];int main(){    int n;    while(~scanf("%d",&n))    {        if(n==0)break;        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {               scanf("%d",&a[i][j]);            }        }        int ans0=0,ans90=0,ans180=0,ans270=0;        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                scanf("%d",&b[i][j]);                if(b[i][j]==a[i][j])                ans0++;            }        }        int x=0,y=0;        for(int j=0;j<n;j++)        {            for(int i=n-1;i>=0;i--)            {                if(a[i][j]==b[x][y])ans90++;                y++;                if(y==n)                {                    y=0;                    x++;                }            }        }        x=0,y=0;        for(int i=n-1;i>=0;i--)        {            for(int j=n-1;j>=0;j--)            {                if(a[i][j]==b[x][y])ans180++;                y++;                if(y==n)                {                    y=0;                    x++;                }            }        }        x=0,y=0;        for(int j=n-1;j>=0;j--)        {            for(int i=0;i<n;i++)            {                if(a[i][j]==b[x][y])ans270++;                y++;                if(y==n)                {                    y=0;                    x++;                }            }        }        printf("%d\n",max(max(ans0,ans90),max(ans180,ans270)));    }    return 0;}