HDU2859 有毒的DP

最新推荐文章于 2019-08-22 16:21:24 发布

NineFailure

最新推荐文章于 2019-08-22 16:21:24 发布

阅读量406

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM练习动态规划

本文链接：https://blog.youkuaiyun.com/NineFailure/article/details/52138668

ACM练习同时被 2 个专栏收录

312 篇文章

订阅专栏

动态规划

88 篇文章

订阅专栏

本文探讨了一道关于寻找最大对称子矩阵的问题，并提供了一种有效的算法解决方案。通过对输入矩阵进行动态规划处理，该算法能够确定任意给定矩阵中最大的左下到右上对称子阵的尺寸。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

卧槽调试了一个下午
竟然是….
初始化有问题！！！！！！！！
有病啊！！！！！！！！！！！！！！！！

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

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<cstdio>
#include<string>
using namespace std;
string zhi[1010];
int dp[1010][1010];
int main()
{
    int n;
    while (cin >> n)
    {
        if (n == 0)break;
        for (int a = 1;a <= n;a++)for (int b = 0;b <= n;b++)dp[a][b] = 1;
        for (int a = 1;a <= n;a++)cin >> zhi[a];
        int sum = 1;
        for (int a = 2;a <= n;a++)
        {
            for (int b = 0;b < n - 1;b++)
            {
                int c;
                for (c = 1;c <= dp[a - 1][b + 1];c++)if (zhi[a - c][b] != zhi[a][b + c])break;
                dp[a][b] = c;
                sum = max(sum, dp[a][b]);
            }
        }
        cout << sum << endl;
    }
    return 0;
}