CODE FESTIVAL 2017 qual A--C - Palindromic Matrix(模拟所有情况，注意细节)

个人心得：其实本来这题是有规律的不过当时已经将整个模拟过程都构思出来了，就打算试试，将每个字符和总和用优先队列

装起来，然后枚举每个点，同时进行位置标志，此时需要多少个点的时候拿出最大的和出来，若不满足就输出No，结果一直卡在三组

数据。比赛完后一想，优先队列虽然用处大，不过当行列存在奇数的时候此时只要2个就可以，而如果你从最大的4个中拿出来，

那么下一层循环中必然有一个位置无法填充，如此就导致了算法的失败。所以后面建立个奇偶判定就好了。

感悟：

多注意思考和细节，从不同的层次看待问题，既可以全面又可以优化所有细节！

题目：

Problem Statement

We have an H-by-W matrix. Let aij be the element at the i-th row from the top and j-th column from the left. In this matrix, each aij is a lowercase English letter.

Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:

  • Every row and column in A' can be read as a palindrome.

Determine whether he can create a matrix satisfying the condition.

Note

A palindrome is a string that reads the same forward and backward. For example, a, aa, abba and abcba are all palindromes, while ab, abab andabcda are not.

Constraints

  • 1≤H,W≤100
  • aij is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

H W
a11a12…a1W
:
aH1aH2…aHW

Output

If Snuke can create a matrix satisfying the condition, print Yes; otherwise, print No.

Sample Input 1

Copy
3 4
aabb
aabb
aacc

Sample Output 1

Copy
Yes

For example, the following matrix satisfies the condition.

abba
acca
abba

Sample Input 2

Copy
2 2
aa
bb

Sample Output 2

Copy
No

It is not possible to create a matrix satisfying the condition, no matter how we rearrange the elements in A.

Sample Input 3

Copy
5 1
t
w
e
e
t

Sample Output 3

Copy
Yes

For example, the following matrix satisfies the condition.

t
e
w
e
t

Sample Input 4

Copy
2 5
abxba
abyba

Sample Output 4

Copy
No

Sample Input 5

Copy
1 1
z

Sample Output 5

Copy
Yes
  1 #include<iostream>
  2 #include<cstring>
  3 #include<string>
  4 #include<cstdio>
  5 #include<vector>
  6 #include<cmath>
  7 #include<stack>
  8 #include<set>
  9 #include<queue>
 10 #include<algorithm>
 11 using namespace std;
 12 #define in 1000000007
 13 int h,w;
 14 char ch[105][105];
 15 int ok=0;
 16 struct mapa
 17 {
 18     char x;
 19     int sum;
 20     mapa(char m,int n)
 21     {
 22         x=m;
 23         sum=n;
 24     }
 25     bool operator <(const mapa &a)const
 26     {
 27         return sum<a.sum;
 28     }
 29 };
 30 int sum[27];
 31 priority_queue<mapa >pq;
 32 int book[105][105];
 33 void flag(int i,int j){
 34       book[i][j]=1;
 35             int t=1;
 36             if(!book[i][w-1-j])
 37             {
 38                 t++;
 39                 book[i][w-1-j]=1;
 40             }
 41             if(!book[h-1-i][j])
 42             {
 43                 t++;
 44                 book[h-1-i][j]=1;
 45             }
 46             if(!book[h-1-i][w-1-j])
 47             {
 48                 t++;
 49                 book[h-1-i][w-1-j]=1;
 50             }
 51            mapa a=pq.top();pq.pop();
 52            if(a.sum<t)
 53            {
 54                ok=1;
 55                return false;
 56            }
 57            a.sum=a.sum-t;
 58            if(a.sum)
 59               pq.push(a);
 60 }
 61 bool dfs()
 62 {
 63     memset(book,0,sizeof(book));
 64     int p=0,q=0;
 65     int i,j;
 66     if(h%2==1) p=1;
 67     if(w%2==1) q=1;
 68     for(i=0;i<h;i++){
 69         if(p&&i==h/2) break;
 70        for(j=0;j<w;j++)
 71        {
 72            if(q&&j==w/2) break;
 73         if(book[i][j]) continue;
 74         else
 75         {
 76            flag(i,j);
 77         }
 78        }
 79        }
 80        if(p)
 81        {
 82            for(int k=0;k<w;k++)
 83            {
 84                if(q&&k==w/2) break;
 85                if(book[i][k]) continue;
 86                book[i][k]=1;
 87             int t=1;
 88             if(!book[i][w-1-k])
 89             {
 90                 t++;
 91                 book[i][w-1-k]=1;
 92             }
 93            mapa a=pq.top();pq.pop();
 94            if(a.sum<t)
 95            {
 96                ok=1;
 97                return false;
 98            }
 99            a.sum=a.sum-t;
100            if(a.sum)
101               pq.push(a);
102         }
103        }
104        if(q)
105        {
106            for(int k=0;k<h;k++)
107            {
108                if(book[k][j]) continue;
109                book[k][j]=1;
110             int t=1;
111             if(!book[h-1-k][j])
112             {
113                 t++;
114                 book[h-1-k][j]=1;
115             }
116            mapa a=pq.top();pq.pop();
117            if(a.sum<t)
118            {
119                ok=1;
120                return false;
121            }
122            a.sum=a.sum-t;
123            if(a.sum)
124               pq.push(a);
125         }
126        }
127        return true;
128 }
129 int main()
130 {
131     cin>>h>>w;
132     for(int i=0;i<h;i++)
133          for(int j=0;j<w;j++)
134              cin>>ch[i][j];
135              memset(sum,0,sizeof(sum));
136     for(int i=0;i<h;i++)
137          for(int j=0;j<w;j++)
138             sum[ch[i][j]-'a']++;
139         for(int i=0;i<27;i++)
140             if(sum[i])
141             {
142                 char t=i+'a';
143                 pq.push(mapa(t,sum[i]));
144             }
145             if(h==1||w==1)
146         {
147             int flag=0;
148             for(int i=0;i<27;i++)
149                 if(sum[i]%2!=0) flag++;
150             if(flag>1) cout<<"No"<<endl;
151             else
152                 cout<<"Yes"<<endl;
153         }
154         else{
155             if(dfs())
156                cout<<"Yes"<<endl;
157                else
158                  cout<<"No"<<endl;
159         }
160 
161     return 0;
162 }

 



 

posted @ 2017-09-24 10:35  余生漫漫浪  阅读(378)  评论(0)  编辑  收藏  举报