第十届ACM省赛-E Plumbing the depth of lake

问题 G: Plumbing the depth of lake

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题目描述

There is a mysterious lake in the north of Tibet. As the sun shines, the surface of the lake is colorful and colorful. The lake was unfathomable in rainy weather.  After the probe,  It has an interesting bottom in that it is full of little hills and valleys. . Scientists wonders how deep the bottom of the lake is.

 

Scientists use the most advanced radar equipment to detect the bottom of the lake. It is the discovery that the deepest part is relatively flat. Thet want to know the largest depth number only if it is verified by the fact that the same depth appears in an adjacent reading.

 

To facilitate computing, scientists have put the lake as M * N grids . The depth  reading of each grid is already known. some readings might be 0-- It's a small island on the lake.

 

Find the greatest depth that appears in at least two 'adjacent'readings (where 'adjacent' means in any of the potentially eight squares that border a square on each of its sides and its diagonals). The lake has at least one pair of positive, adjacent readings.

输入

The first line of the input contains one integers T, which is the nember of  test cases (1<=T<=5).  Each test case specifies:

 

* Line 1:      Two space-separated integers: M and N   (1 ≤ M,  N ≤ 50)

* Lines 2..M+1: Line i+1 contains N space-separated integers that  represent the depth of the lake across row i: Dij    (0 <= Dij <=1,000,000);

输出

For each test case generate a single line:  a single integer that is the depth of the lake determined.

样例输入

1

4 3

0 1 0

1 2 0

1 5 1

2 3 4

样例输出

1

思路：

求 最深深度   当一片区域存在至少紧挨着的两个相同深度  达到要求  求出 满足条件的最深深度

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int x[8]={-1,-1,-1,0,0,1,1,1}; 
int y[8]={-1,0,1,-1,1,-1,0,1};
int main()
{
    int t,i,j;
    int a[100][100];
    cin>>t;
    while(t--)
    {
        int max=-1;  //初值附为-1
        int m,n;
        memset(a,-1,sizeof(a));
        cin>>m>>n;
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                cin>>a[i][j];
            }
        }
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                if(max>=a[i][j])
                {
                    continue;
                }
                for(int k=0;k<8;k++)
                {
                    if(a[i][j]==a[i+x[k]][j+y[k]]) //八个方向
                    {
                        max=a[i][j];
                        break;
                    }
                }
            }
        }
        cout<<max<<endl;
    }
}