HDU 2859—Phalanx（DP）

Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

大意：

给定一个矩阵，求最大对称矩阵

思路：

一开始怎么想也觉得不会有状态转移的情况，即使有也会复杂度过高

后来看了题解才想通，其实和我一开始的想的有些相似，因为给的时间比较长，n^3的复杂度也会死可以接受的

试想一个n阶的对称阵如何变成n+1的对称阵？

在它的两个底边追加对称的元素即可，对角线元素添加任意元素即可

于是，我们从一个点向他的上与右边推进，直至不匹配

将匹配的个数与dp[i-1][j+1]比较，

如果匹配量大于右上角记录下来的矩阵大小，就是右上角的数值+1，否则就是这个匹配量。

代码：

#include<bits/stdc++.h>

using namespace std;

const int MAXN=1300;

char m[MAXN][MAXN];

int dp[MAXN][MAXN];

int main()

{

    //freopen("data.in","r",stdin);

    int n;

    while(~scanf("%d",&n)&&n){

        getchar();

        for(int i=0;i<n;i++){

            for(int j=0;j<n;j++){

                scanf("%c",&m[i][j]);

            }

            getchar();

        }

//        for(int i=0;i<n;i++){

//            for(int j=0;j<n;j++){

//                cout<<m[i][j];

//            }

//        }

//        cout<<endl;

        memset(dp,0,sizeof(dp));

        int res=1;

        for(int i=0;i<n;i++){

            for(int j=n-1;j>=0;j--){

                //cout<<i<<"\t"<<j<<"\t";

                //cout<<1111111<<endl;

                if(i==0||j==n-1){

                    dp[i][j]=1;

                    //cout<<"is "<<dp[i][j]<<endl;

                    continue;

                }

                int x=i,y=j;

                while(x>=0&&y<=n-1&&m[x][j]==m[i][y]){

                    x--;

                    y++;

                }

                y=y-j;

                if(y>dp[i-1][j+1]){

                    dp[i][j]=dp[i-1][j+1]+1;

                }

                else{

                    dp[i][j]=y;

                }

                //cout<<"is  "<<dp[i][j]<<endl;

                res=max(res,dp[i][j]);

            }

        }

        printf("%d\n",res);

    }

}

巴特西

HDU 2859—Phalanx（DP）

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