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:求出从左下角到右上角的对角线为对称线的最大的矩阵。
思路:用dp[i][j]表示在点(i,j)起向上和向右的线上相对称的点多于dp[i-1][j+1],则dp[i][j]=dp[i-1][j+1]+1.
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
int n;
char p[1010][1010];
int dp[1010][1010];
while(scanf("%d",&n)!=EOF&&n!=0){
for(int i=0;i<n;i++){
scanf("%s",p[i]);
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
dp[i][j]=1;
}
}
int i,j,t1,t2,ans,ansmax=0;
for(i=0;i<n;i++){
for(j=0;j<n;j++){
if(i==0||j==n-1) continue;//边界判断
t1=i;t2=j;
ans=0;
while(t1>=0&&t2<n&&p[t1][j]==p[i][t2]){
t1--;t2++;ans++;
}
if(ans>dp[i-1][j+1]){
dp[i][j]=dp[i-1][j+1]+1;
}
else dp[i][j]=ans;
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
ansmax=max(ansmax,dp[i][j]);
}
}
cout<<ansmax<<endl;
}
return 0;
}