hdu 4331 DP

看了那么多 ，暴力就能过啊卧槽！

这题开始我的题意也理解错了。。。以为是求实心矩阵，最后发现只要是边上全部是1就行。

那么只需要四个数组，记录每一点能够 向上、向下、向左、向右所能延伸的最长的距离

然后暴力枚举所有的点就行。。。。

AC代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

int lf[1100][1100], rt[1100][1100], up[1100][1100], down[1100][1100];
int num[1100][1100];
int N, T, ans;

void init(){
    memset( lf, 0, sizeof( lf ) );
    memset( rt, 0, sizeof( rt ) );
    memset( up, 0, sizeof( up ) );
    memset( down, 0, sizeof( down ) );

    for( int i = 1; i <= N; i++ ){
        for( int j = 1; j <= N; j++ ){
            if( !num[i][j] )    continue;
            if( num[i][j] ) lf[i][j] = 1;
            lf[i][j] += lf[i][j-1];
        }
        for( int j = N; j >= 1; j-- ){
            if( !num[i][j] )    continue;
            if( num[i][j] ) rt[i][j] = 1;
            rt[i][j] += rt[i][j+1];
        }
    }

    for( int j = 1; j <= N; j++ ){
        for( int i = 1; i <= N; i++ ){
            if( !num[i][j] )    continue;
            if( num[i][j] ) up[i][j] = 1;
            up[i][j] += up[i-1][j];
        }
        for( int i = N; i >= 1; i-- ){
            if( !num[i][j] )    continue;
            if( num[i][j] ) down[i][j] = 1;
            down[i][j] += down[i+1][j];
        }
    }
}

int solve(){

    if( ans == N * N ){//注意这种情况
        return ( 2 * N + 1 ) * ( N + 1 ) * N / 6;;
    }

    for( int i = 1; i <= N; i++ ){
        for( int j = 1; j <= N; j++ ){
            if( num[i][j] == 0 ) continue;
            int t = min( rt[i][j], down[i][j] );
            for( int k = 2; k <= t; k++ ){
                int x = i + k - 1;
                int y = j + k - 1;
                if( lf[x][y] >= k && up[x][y] >= k )  ans++;
            }
        }
    }

    return ans;
}

int main(){
    int Case = 1;
    scanf( "%d", &T );
    while( T-- ){
        scanf( "%d", &N );
        ans = 0;
        for( int i = 1; i <= N; i++ ){
            for( int j = 1; j <= N; j++ ){
                scanf( "%d", &num[i][j] );
                if( num[i][j] ){
                    ans++;
                }
            }
        }
        init();
        printf( "Case %d: %d\n", Case++, solve() );
    }
    return 0;
}