king's trouble II SCU - 4488

Time Limit: 1000 MS Memory Limit: 131072 K

Description

Long time ago, a king occupied a vast territory.
Now there is a problem that he worried that he want to choose a largest square of his territory to build a palace.
Can you help him?

For simplicity, we use a matrix to represent the territory as follows:
0 0 0 0 0
0 1 0 1 0
1 1 1 1 0
0 1 1 0 0
0 0 1 0 0

Every single number in the matrix represents a piece of land, which is a 1*1 square
1 represents that this land has been occupied
0 represents not

Obviously the side-length of the largest square is 2

Input

The first line of the input contains a single integer t (1≤t≤5) — the number of cases.
For each case
The first line has two integers N and M representing the length and width of the matrix
Then M lines follows to describe the matrix
1≤N,M≤1000

Output

For each case output the the side-length of the largest square

Sample Input

2
5 5
0 0 0 0 0
0 1 0 1 0
1 1 0 1 0
0 1 1 0 0
0 0 1 0 0
5 5
0 0 0 0 0
0 1 0 1 0
1 1 1 1 0
0 1 1 0 0
0 0 1 0 0

Sample Output

1
2

#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

const int INF=1e9+7;
const int maxn=1010;
typedef long long ll;


//单调栈
//先做 poj 2559后再看单调栈可能会好理解一点
int a[1010][1010];
struct node
{
    int idx,num;//下标和延伸值
};
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                scanf("%d",&a[i][j]);
                if(a[i][j])
                    a[i][j]+=a[i-1][j];
            }
            a[i][m+1]=-1;
        }
        int ans=-1;
        for(int i=1; i<=n; i++)
        {
            int top=0;
            node vis[1010];
            for(int j=1; j<=m+1; j++)
            {
                if(top==0)
                {
                    vis[top].idx=j,vis[top++].num=1;

                }
                else
                {
                    printf("top=%d\n",top);
                    int v=vis[top-1].idx,cont=0;
                    while(top>0&&a[i][v]>=a[i][j])
                    {
                        cont+=vis[top-1].num;
                        if(cont>=a[i][v])
                            ans=max(ans,a[i][v]);
                        top--;
                        v=vis[top-1].idx;
                    }
                    vis[top].idx=j,vis[top++].num=cont+1;
                    printf("vis[top].idx=%d vis[top].num=%d\n",vis[top-1].idx,vis[top-1].num);
                }
            }
        }
        printf("%d\n",ans);
    }
}


//暴力
//先把每一列累加,类似于条形图,然后从每一行的开头开始遍历到每一行结束,
//(不过是因为这一题的后台数据水,
//暴力才能过,所以如果数据范围太大不推荐暴力)
int mp[maxn][maxn];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m,x;
        memset(mp,0,sizeof(mp));
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                scanf("%d",&x);
                if(x&&mp[i-1][j])
                    mp[i][j]=mp[i-1][j]+1;
                else
                    mp[i][j]=x;
            }
        }
        int maxx=0;
        int k=1;//代表此时最大的面积
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                if(mp[i][j]>=k)
                {
                    int p=0;
                    for(; mp[i][p+j]>=k; p++);//每一行从头遍历到尾
                    if(p>=k)
                    {
                        maxx=k;
                        k++;
                    }
                }
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}



//动规
//以每一个点作为正方形的右上角,
//每次取四个角的最小值加一即为此时的最大面积(因为本题规定的每个矩形面积为1)
//自己画图可以理解的
int a[maxn][maxn];
int dp[maxn][maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        memset(dp,0,sizeof(dp));
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                scanf("%d",&a[i][j]);
        int ans=0;
        for(int i=1; i<n; i++)
        {
            for(int j=1; j<m; j++)
            {
                if(a[i][j]==1)
                {
                    dp[i][j]=min(dp[i][j-1],min(dp[i-1][j-1],dp[i-1][j]))+1;
                }
                ans=max(ans,dp[i][j]);
            }
        }
        printf("%d\n",ans);
    }
}