ACM DPPhalanx

xiaoxiao2021-02-28  3

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;}

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