标签：atcoder Sample SS TT shapes ..... beginner218 Input Copy
一、题目大意
题目链接：https://atcoder.jp/contests/abc218/tasks/abc218_c
Problem Statement
We have two figures SS and TT on a two-dimensional grid with square cells.
SS lies within a grid with NN rows and NN columns, and consists of the cells where S_{i,j}Si,j is #.
TT lies within the same grid with NN rows and NN columns, and consists of the cells where T_{i,j}Ti,j is #.
Determine whether it is possible to exactly match SS and TT by 9090-degree rotations and translations.
Constraints
- 1 \leq N \leq 2001≤N≤200
- Each of SS and TT consists of
#and.. - Each of SS and TT contains at least one
#.
Input
Input is given from Standard Input in the following format:
NN
S_{1,1}S_{1,2}\ldots S_{1,N}S1,1S1,2…S1,N
\vdots⋮
S_{N,1}S_{N,2}\ldots S_{N,N}SN,1SN,2…SN,N
T_{1,1}T_{1,2}\ldots T_{1,N}T1,1T1,2…T1,N
\vdots⋮
T_{N,1}T_{N,2}\ldots T_{N,N}TN,1TN,2…TN,N
Output
Print Yes if it is possible to exactly match SS and TT by 9090-degree rotations and translations, and No otherwise.
Sample Input 1 Copy
Copy5 ..... ..#.. .###. ..... ..... ..... ..... ....# ...## ....#
Sample Output 1 Copy
CopyYes
We can match SS to TT by rotating it 9090-degrees counter-clockwise and translating it.
Sample Input 2 Copy
Copy5 ##### ##..# #..## ##### ..... ##### #..## ##..# ##### .....
Sample Output 2 Copy
CopyNo
It is impossible to match them by 9090-degree rotations and translations.
Sample Input 3 Copy
Copy4 #... ..#. ..#. .... #... #... ..#. ....
Sample Output 3 Copy
CopyYes
Each of SS and TT may not be connected.
Sample Input 4 Copy
Copy4 #... .##. ..#. .... ##.. #... ..#. ....
Sample Output 4 Copy
CopyNo
Note that it is not allowed to rotate or translate just a part of a figure; it is only allowed to rotate or translate a whole figure.
大意就是给两个矩阵，看里面的'#'能否通过旋转、平移重合
二、ac代码
#include<bits/stdc++.h>
using namespace std;
struct node{
int x,y;
};
map<int,node>mp;
bool comp(const node &a,const node &b)
{
if (a.x==b.x) return a.y<b.y;
else return a.x<b.x;
}
int main(void)
{
string s;
vector<node>vec;
ios::sync_with_stdio(0);
int n,cnt=0,flag=1;
cin>>n;
for (int i=1;i<=n;i++)
{
cin>>s;
for (int j=0;j<n;j++)
{
if (s[j]=='#')
{
node t;
t.x=i,t.y=j+1;
mp[cnt++]=t; //序号与坐标
}
}
}
for (int i=1;i<=n;i++)
{
cin>>s;
for (int j=0;j<n;j++)
{
if (s[j]=='#')
{
node t;
t.x=i,t.y=j+1;
vec.push_back(t);
}
}
}
if (cnt!=vec.size()) //个数不相等，肯定不可能重合
{
cout<<"No";
return 0;
}
for (int i=1;i<=4;i++)
{
flag=1;
for (int j=0;j<vec.size();j++) //更新旋转后的坐标
{
swap(vec[j].x,vec[j].y);
vec[j].y=n-vec[j].y+1;
}
sort(vec.begin(),vec.end(),comp); //按照从左到右，从上到下的顺序排序
int row=vec[0].x-mp[0].x; //判断旋转后的第一个坐标和要比较的第一个坐标的行的关系
int column=vec[0].y-mp[0].y; //判断旋转后的第一个坐标和要比较的第一个坐标的列的关系
for (int j=1;j<cnt;j++)
{
if (mp[j].x+row==vec[j].x&&mp[j].y+column==vec[j].y) continue;
else
{
flag=0;
break;
}
}
if (flag) break;
}
if (flag) cout<<"Yes";
else cout<<"No";
return 0;
}
思路就是每次将第二个矩阵旋转90°，依次判断每个点的距离是不是完全一样
..... ..#.. .###. ..... ..... ..... ..... ....# ...## ....#
以这个样例为例，如果能旋转平移重合的话，每个点相对应的关系是一样的，类似相似三角形
..#.. .###. ..... ..... .....
.....
.....
.....
.#...
###..
因此，在开始的map中已经保存好了上面矩阵的信息，下面的矩阵每次旋转完sort排序，得到的就是从左到右从上到下依次的点
peace
标签：atcoder,Sample,SS,TT,shapes,.....,beginner218,Input,Copy 来源： https://www.cnblogs.com/bcyyds/p/15260047.html
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