Q - Phalanx
Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 2859
Description
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
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
0
Sample Output
3
3
搜索每一个位置作为对角线的情况下能够到达的最大值
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1005;
int dp[maxn][maxn];
char s[maxn][maxn];
int N,ans;
void work(int x,int y)//枚举每一个点的上方与右边能够到的最大值
{
if(x==0||y==N-1){
dp[x][y]=1;
return ;
}
int i=x,j=y;
while(i>=0&&j<N&&s[i][y]==s[x][j]){
i--;j++;
}
int len=x-i;
if(len>=dp[x-1][y+1]+1)dp[x][y]=dp[x-1][y+1]+1;
else dp[x][y]=len;
ans=max(ans,dp[x][y]);
}
int main()
{
while(scanf("%d",&N)&&N){
for(int i=0;i<N;i++){
scanf("%s",s[i]);
}
ans=1;
for(int i=0;i<N;i++)
for(int j=0;j<N;j++){
work(i,j);
}
printf("%d\n",ans);
}
return 0;
}
Q - Phalanx HDU 2859 ( dp )
原文：http://blog.csdn.net/u013167299/article/details/45788735
