ZOJ 3838 Infusion Altar(数学啊 模拟啊 这么挫的代码你见过吗?)
2014-11-30 18:13
453 查看
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5412
Bob is recently playing a game called Minecraft, especially a mod called Thaumcraft. It is a mod of magic.
Usually, Bob has Obsessions with Symmetry while playing Minecraft. This obsession is useless in the gameplay generally. However, in Thaumcraft, the infusion altar requires
symmetry to keep it stable.
Bob built an infusion altar in his secret chamber, but it was not so symmetrical. After some explosions, Bob decided to fix the infusion altar to make it symmetrical.
![](https://oscdn.geek-share.com/Uploads/Images/Content/202004/10/516b0e445e2613bd68979fcd546dad53.jpg)
You will be given the map of Bob's secret chamber. It is of size n*n(n is an odd number), the infusion altar is always at the center of his secret chamber.
The following picture is a typical map. The 3*3 square in the center is the Infusion Altar, it is a multi-block structure. Here, '#' means Runic Matrix, 'o' means Arcane Pedestal, '.' means an empty place, 'a'-'z' means occult paraphernalia(like skulls, crystals
and candles) Bob placed around the Infusion Altar. There will not be characters other than 'a'-'z', '.', '#'.
Now, the question is that at least how many blocks need to be changed to make the whole map symmetrical. Here, being symmetrical means having all four axes of symmetry
for a square. Also, you can change any character on the map to any other character.
For each case, The first line contains an integer n ( 3 ≤ n ≤ 99, and n is an odd number)
For the next n lines, each line contains n characters showing the map.
It is guaranteed that the Infusion Altar is at the center of the map.
It is guaranteed that only 'a'-'z' and '.' will appear out of the Infusion Altar.
In second sample, Bob will change his secret chamber to the following map.
Author: ZHU, Jiale; GONG, Yuan
Source: ZOJ Monthly, November 2014
题意:
求最少的操作次数使图形对称!
PS:
![](https://oscdn.geek-share.com/Uploads/Images/Content/201910/01/5bb18535e0dd0414e2084256df69a6c8)
1、先分别找出要使四条对称轴对称的最少操作次数,(显然只需要分别找出其中一条红色线和其中一条白色线的一半使八个部分一样即可);
2、再枚举1、2、3……号格子分别和其余七个格子相同最少所需的次数!
![](https://oscdn.geek-share.com/Uploads/Images/Content/201910/01/0211d519911aa189b1897da2d29b0991)
代码如下:(这么挫的代码你见过吗?)
Bob is recently playing a game called Minecraft, especially a mod called Thaumcraft. It is a mod of magic.
Usually, Bob has Obsessions with Symmetry while playing Minecraft. This obsession is useless in the gameplay generally. However, in Thaumcraft, the infusion altar requires
symmetry to keep it stable.
Bob built an infusion altar in his secret chamber, but it was not so symmetrical. After some explosions, Bob decided to fix the infusion altar to make it symmetrical.
![](https://oscdn.geek-share.com/Uploads/Images/Content/202004/10/516b0e445e2613bd68979fcd546dad53.jpg)
You will be given the map of Bob's secret chamber. It is of size n*n(n is an odd number), the infusion altar is always at the center of his secret chamber.
The following picture is a typical map. The 3*3 square in the center is the Infusion Altar, it is a multi-block structure. Here, '#' means Runic Matrix, 'o' means Arcane Pedestal, '.' means an empty place, 'a'-'z' means occult paraphernalia(like skulls, crystals
and candles) Bob placed around the Infusion Altar. There will not be characters other than 'a'-'z', '.', '#'.
.aab. bo.ob b.#.a bo.ob bbab.
Now, the question is that at least how many blocks need to be changed to make the whole map symmetrical. Here, being symmetrical means having all four axes of symmetry
for a square. Also, you can change any character on the map to any other character.
Input
There are multiple cases. The first line contains one integer T which is the number of test cases.For each case, The first line contains an integer n ( 3 ≤ n ≤ 99, and n is an odd number)
For the next n lines, each line contains n characters showing the map.
It is guaranteed that the Infusion Altar is at the center of the map.
It is guaranteed that only 'a'-'z' and '.' will appear out of the Infusion Altar.
Output
One integer for each test case which is the least number of blocks that should be changed.Sample Input
3
3
o.o
.#.
o.o
5
.aab. bo.ob b.#.a bo.ob bbab.5
aabba
ao.oa
a.#.a
ao.oa
aaaaa
Sample Output
0 3 2
Hint
The first sample is a standard Infusion Altar.In second sample, Bob will change his secret chamber to the following map.
.bab. bo.ob a.#.a bo.ob .bab.
Author: ZHU, Jiale; GONG, Yuan
Source: ZOJ Monthly, November 2014
题意:
求最少的操作次数使图形对称!
PS:
1、先分别找出要使四条对称轴对称的最少操作次数,(显然只需要分别找出其中一条红色线和其中一条白色线的一半使八个部分一样即可);
2、再枚举1、2、3……号格子分别和其余七个格子相同最少所需的次数!
代码如下:(这么挫的代码你见过吗?)
#include <cstdio>
#include <cstring>
#include <cmath>
int main()
{
int t;
int n;
char s[117][117];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i = 0; i < n; i++)
{
scanf("%s",s[i]);
}
int m = n/2;
int sum = 0;
int minn;
char tt;
for(int i = 0; i < m; i++)//中线
{
minn = 4;
tt = s[i][m];//1
int k1 = 0;
if(tt != s[m][i])
k1++;
if(tt != s[n-1-i][m])
k1++;
if(tt != s[m][n-1-i])
k1++;
if(minn > k1)
minn = k1;
tt = s[m][i];//2
int k2 = 0;
if(tt != s[i][m])
k2++;
if(tt != s[n-1-i][m])
k2++;
if(tt != s[m][n-1-i])
k2++;
if(minn > k2)
minn = k2;
tt = s[n-1-i][m];//3
int k3 = 0;
if(tt != s[m][i])
k3++;
if(tt != s[i][m])
k3++;
if(tt != s[m][n-1-i])
k3++;
if(minn > k3)
minn = k3;
tt = s[m][n-1-i];//4
int k4 = 0;
if(tt != s[m][i])
k4++;
if(tt != s[n-1-i][m])
k4++;
if(tt != s[i][m])
k4++;
if(minn > k4)
minn = k4;
sum+=minn;
}
for(int i = 0; i < m; i++)//主副对角线
{
minn = 4;
tt = s[i][i];//1
int k1 = 0;
if(tt != s[n-1-i][n-1-i])
k1++;
if(tt != s[n-1-i][i])
k1++;
if(tt != s[i][n-1-i])
k1++;
if(minn > k1)
minn = k1;
tt = s[n-1-i][n-1-i];//2
int k2 = 0;
if(tt != s[i][i])
k2++;
if(tt != s[n-1-i][i])
k2++;
if(tt != s[i][n-1-i])
k2++;
if(minn > k2)
minn = k2;
tt = s[n-1-i][i];//3
int k3 = 0;
if(tt != s[n-1-i][n-1-i])
k3++;
if(tt != s[i][i])
k3++;
if(tt != s[i][n-1-i])
k3++;
if(minn > k3)
minn = k3;
tt = s[i][n-1-i];//4
int k4 = 0;
if(tt != s[n-1-i][n-1-i])
k4++;
if(tt != s[n-1-i][i])
k4++;
if(tt != s[i][i])
k4++;
if(minn > k4)
minn = k4;
sum+=minn;
}
//printf("sum::%d\n",sum);
//int cen = ((n/2)*(n/2)-(n/2))/2;
int cen = n/2;
//printf("cen::%d\n",cen);
for(int i = 0; i < cen-1; i++)
{
int cc = 0;
for(int j = i+1; j < cen; j++)
{
minn = 8;
tt = s[i][j];//1
//printf("s[i][j]:%c\n",s[i][j]);
cc = 0;
if(tt != s[j][i])
cc++;
//printf("s[i][j]:%c\n",s[j][i]);
if(tt != s[i][n-1-j])
cc++;
//printf("s[i][n-1-j]:%c\n",s[i][n-1-j]);
if(tt != s[n-1-j][i])
cc++;
//printf("s[n-1-j][i]:%c\n",s[n-1-j][i]);
if(tt != s[j][n-1-i])
cc++;
//printf("s[j][n-1-i]:%c\n",s[j][n-1-i]);
if(tt != s[n-1-i][j])
cc++;
//printf("s[n-1-i][j]:%c\n",s[n-1-i][j]);
if(tt != s[n-1-j][n-1-i])
cc++;
//printf("s[n-1-j][n-1-i]:%c\n",s[n-1-j][n-1-i]);
if(tt != s[n-1-i][n-1-j])
cc++;
//printf("s[n-1-i][n-1-j]:%c\n",s[n-1-i][n-1-j]);
if(cc < minn)
minn = cc;
tt = s[j][i];//2
cc = 0;
if(tt != s[i][j])
cc++;
if(tt != s[i][n-1-j])
cc++;
if(tt != s[n-1-j][i])
cc++;
if(tt != s[j][n-1-i])
cc++;
if(tt != s[n-1-i][j])
cc++;
if(tt != s[n-1-j][n-1-i])
cc++;
if(tt != s[n-1-i][n-1-j])
cc++;
if(cc < minn)
minn = cc;
tt = s[i][n-1-j];//3
cc = 0;
if(tt != s[j][i])
cc++;
if(tt != s[i][j])
cc++;
if(tt != s[n-1-j][i])
cc++;
if(tt != s[j][n-1-i])
cc++;
if(tt != s[n-1-i][j])
cc++;
if(tt != s[n-1-j][n-1-i])
cc++;
if(tt != s[n-1-i][n-1-j])
cc++;
if(cc < minn)
minn = cc;
tt = s[n-1-j][i];//4
cc = 0;
if(tt != s[j][i])
cc++;
if(tt != s[i][n-1-j])
cc++;
if(tt != s[i][j])
cc++;
if(tt != s[j][n-1-i])
cc++;
if(tt != s[n-1-i][j])
cc++;
if(tt != s[n-1-j][n-1-i])
cc++;
if(tt != s[n-1-i][n-1-j])
cc++;
if(cc < minn)
minn = cc;
tt = s[j][n-1-i];//5
cc = 0;
if(tt != s[j][i])
cc++;
if(tt != s[i][n-1-j])
cc++;
if(tt != s[n-1-j][i])
cc++;
if(tt != s[i][j])
cc++;
if(tt != s[n-1-i][j])
cc++;
if(tt != s[n-1-j][n-1-i])
cc++;
if(tt != s[n-1-i][n-1-j])
cc++;
if(cc < minn)
minn = cc;
tt = s[n-1-i][j];//6
cc = 0;
if(tt != s[j][i])
cc++;
if(tt != s[i][n-1-j])
cc++;
if(tt != s[n-1-j][i])
cc++;
if(tt != s[j][n-1-i])
cc++;
if(tt != s[i][j])
cc++;
if(tt != s[n-1-j][n-1-i])
cc++;
if(tt != s[n-1-i][n-1-j])
cc++;
if(cc < minn)
minn = cc;
tt = s[n-1-j][n-1-i];//7
cc = 0;
if(tt != s[j][i])
cc++;
if(tt != s[i][n-1-j])
cc++;
if(tt != s[n-1-j][i])
cc++;
if(tt != s[j][n-1-i])
cc++;
if(tt != s[n-1-i][j])
cc++;
if(tt != s[i][j])
cc++;
if(tt != s[n-1-i][n-1-j])
cc++;
if(cc < minn)
minn = cc;
tt = s[n-1-i][n-1-j];//8
cc = 0;
if(tt != s[j][i])
cc++;
if(tt != s[i][n-1-j])
cc++;
if(tt != s[n-1-j][i])
cc++;
if(tt != s[j][n-1-i])
cc++;
if(tt != s[n-1-i][j])
cc++;
if(tt != s[n-1-j][n-1-i])
cc++;
if(tt != s[i][j])
cc++;
if(cc < minn)
minn = cc;
sum+=minn;
}
}
printf("%d\n",sum);
}
return 0;
}
/*
99
3
o.o
.#.
o.o
5
.aab. bo.ob b.#.a bo.ob bbab.5
aabba
ao.oa
a.#.a
ao.oa
aaaaa
5
5
.1a2.
3o.o4
b.#.a
5o.o6
b7a8.
*/
相关文章推荐
- 【水题】 ZOJ 3838 Infusion Altar
- 这么详细的分类 PHP源代码网站你见过没有? PHP中的每一个方法,每一个操作,每一个头文件,每一个函数,每一个类.每一个属性,每一个事件都有相应的范例代码
- 这么详细的分类 JSP源代码网站你见过没有? JSP中的每一个方法,每一个操作,每一个函数,每一个类.每一个属性,每一个事件都有相应的范例代码
- [数学](JZOJ)【普及模拟】数列
- 这么详细的分类 VB/VB.NET源代码网站你见过没有? VB/VB.NET中的每一个方法,每一个操作,每一个函数,每一个类.每一个属性,每一个事件都有相应的范例代码
- ZOJ-3327-Friend Number【7th浙江省赛】【模拟】【数学】
- zoj 3879 Capture the Flag(模拟 数学)
- 这么详细的分类 Java源代码网站你见过没有? Java中的每一个方法,每一个操作,每一个函数,每一个类.每一个属性,每一个事件都有相应的范例代码
- 这么详细的分类 C# / C Sharp源代码网站你见过没有? C# / C Sharp中的每一个方法,每一个操作,每一个函数,每一个类.每一个属性,每一个事件都有相应的范例代码
- 这么详细的分类 JavaScript 源代码网站你见过没有? JavaScript 中的每一个方法,每一个操作,每一个函数,每一个类.每一个属性,每一个事件都有相应的范例代码
- zoj--3935--2016(暴力模拟)(数学)
- 这么详细的分类 C++ 源代码网站你见过没有? C++ 中的每一个方法,每一个操作,每一个头文件,每一个函数,每一个类.每一个属性,每一个事件都有相应的范例代码
- 这么详细的分类 PostgreSQL源代码网站你见过没有? PostgreSQL中的每一个方法,每一个操作,每一个函数,每一个类.每一个属性,每一个事件都有相应的范例代码
- 原来,数学也可以这么美的-采用matlab画爱心(附带 MatLab 代码)
- zoj1494 暴力模拟 简单数学问题
- HDU 1988 & ZOJ 2991 Flipping Burned Pancakes(数学啊+模拟)
- [数学](JZOJ)【普及模拟】单元格
- 这么详细的分类 MySQL源代码网站你见过没有? MySQL中的每一个方法,每一个操作,每一个函数,每一个类.每一个属性,每一个事件都有相应的范例代码
- zoj 2106 Tick and Tick(比较好的数学题目,代码特麻烦,注意精度)
- 这么详细的分类 VC/C++源代码网站你见过没有? VC/C++中的每一个方法,每一个操作,每一个头文件,每一个函数,每一个类.每一个属性,每一个事件都有相应的范例代码