HDU2859 有毒的DP

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;}