Phalanx
Today is army day, but the servicemen are busy with the phalanx for the celebration of the 60th anniversary of the PRC.
A phalanx is a matrix of size n*n, each element is a character (a~z or A~Z), standing for the military branch of the servicemen on that position.
For some special requirement it has to find out the size of the max symmetrical sub-array. And with no doubt, the Central Military Committee gave this task to ALPCs.
A symmetrical matrix is such a matrix that it is symmetrical by the “left-down to right-up” line. The element on the corresponding place should be the same. For example, here is a 3*3 symmetrical matrix:
cbx
cpb
zcc
A phalanx is a matrix of size n*n, each element is a character (a~z or A~Z), standing for the military branch of the servicemen on that position.
For some special requirement it has to find out the size of the max symmetrical sub-array. And with no doubt, the Central Military Committee gave this task to ALPCs.
A symmetrical matrix is such a matrix that it is symmetrical by the “left-down to right-up” line. The element on the corresponding place should be the same. For example, here is a 3*3 symmetrical matrix:
cbx
cpb
zcc
Input There are several test cases in the input file. Each case starts with an integer n (0<n<=1000), followed by n lines which has n character. There won’t be any blank spaces between characters or the end of line. The input file is ended with a 0. Output Each test case output one line, the size of the maximum symmetrical sub- matrix.
Sample Input
3 abx cyb zca 4 zaba cbab abbc cacq 0Sample Output
3 3
题意:
给你一个n*n的矩阵,求矩阵的最大反对角线对称的子矩阵的长度。
分析:
由于长度为k的对称矩阵一定包含了一个长度为k-1的对称矩阵,即长度为k的矩阵可以有长度为k-1的矩阵推出来。所以形成了子问题。
我们如何表示一个状态呢?
设dp[i][j] = k,表示左下角坐标为(i,j)长度为k的对称矩阵。
然后我们从“极限状态”出发,当长度为1时,矩阵都为对称矩阵。我们从右上角往左下角方向枚举 ,dp[i][j] 可由dp[i-1][j+1] 推出来。所以直接dp即可。
code:
#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxn = 1005; char a[maxn][maxn]; int dp[maxn][maxn]; int main(){ int n,ans; while(scanf("%d",&n) != EOF && n){ ans = 1; for(int i = 0; i < n; i++){ scanf("%s",a[i]); } //从右上角开始枚举 for(int i = 0; i < n; i++){ for(int j = n-1; j >= 0; j--){ dp[i][j] = 1; if(i == 0 || j == n-1) continue;//在边缘长度只能为1 int k = dp[i-1][j+1]; for(int s = 1; s <= k; s++){ //因为长度为k+1的对称矩阵一定包含了长度为k的对称矩阵,所以判断时只需要判断最外圈是否对称即可 if(a[i-s][j] == a[i][j+s]) dp[i][j]++; else break; } ans = max(ans,dp[i][j]); } } printf("%d\n",ans); } return 0; }