题目:
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 InputThere 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.OutputEach 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大意：给你一个字母矩阵，求出里面的最大反对角矩阵
思路：一个大矩阵对应多个小矩阵的，相当于一个问题对应多个前问题。很显然是一个DP问题，那我们就要找到大矩阵跟小矩阵的关系，沿着大矩阵的对角线，可以发现大矩阵是由小矩阵组成的
用dp[i][j]表示左下角坐标为i，j的反对角矩阵，当然，从右上角也可以。写出dp[i][j]与dp[i-1][j+1]的关系。
代码如下：
#include<stdio.h>#include<algorithm>using namespace std;int dp[1111][1111],n;char a[1111][1111];int main(){ int b,ans; while(scanf("%d",&n)!=EOF) { if(n==0) break; ans=1;//不能是0，因为只有一个字母时 在下面的循环中ans的值是不会改变的； for(int i=0;i<n;i++) scanf("%s",a[i]); 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; b=dp[i-1][j+1]; for(int m=1;m<=b;m++) { if(a[i-m][j]==a[i][j+m]) dp[i][j]++; else break;//有不满足的即使后面满足也不是对应大小的对称矩阵了 } ans=max(ans,dp[i][j]); } } printf("%d\n",ans); } return 0;}
