VJ组队赛->House Building（5538）

额..... 题目太长我就不贴了，就是给你一个二维数组，a[i][j]代表（i,j）位置木块的高度，然后让你求 ｉ x j 这个范围内所有木块的表面积之和，具体如下图

Input

The first line contains an integer T indicating the total number of test cases.
First line of each test case is a line with two integers n,m.
The n lines that follow describe the array of Nyanko-san's blueprint, the i-th of these lines has m integers ci,1,ci,2,...,ci,m, separated by a single space.

1≤T≤50
1≤n,m≤50
0≤ci,j≤1000

 

Output

For each test case, please output the number of glass units you need to collect to meet Nyanko-san's requirement in one line.

 

Sample Input

23 31 0 03 1 21 1 03 31 0 10 0 01 0 1

Sample Output

3020

其实这道题，，，比赛的时候看了一段时间，结果没想到怎么做，不过很快就被队长一发解决了～我知道要扫描每个坐标的上下左右和顶部，不过我真的没想到，,原来扫描完直接加上高度差就是面积啊(@_@)，第一次这样求表面积... 一道水题吧... QAQ

#include <iostream>#include <cstring>using namespace std;const int N = 1e3 + 5;int main(){int a[N][N];int t,n,m;cin >> t;while(t--){cin >> n >> m;memset(a,0,sizeof(a));int sum = 0;int i,j;for(i = 1; i <= n; i++)for(j = 1; j <= m; j++){cin >> a[i][j];if(a[i][j] > 0)            //顶部sum ++;}for(i = 1; i <= n; i++)for(j = 1; j <= m; j++){if(a[i][j] > a[i][j-1])    //左边sum += a[i][j]-a[i][j-1];}for(i = 1; i <= n; i++)for(j = 1; j <= m; j++){if(a[i][j] > a[i][j+1])    //右边sum += a[i][j]-a[i][j+1];}for(i = 1; i <= n; i++)for(j = 1; j <= m; j++){if(a[i][j] > a[i-1][j])    //上边sum += a[i][j]-a[i-1][j];}for(i = 1; i <= n; i++)for(j = 1; j <= m; j++){if(a[i][j] > a[i+1][j])    //下边sum += a[i][j]-a[i+1][j];}cout << sum << "\n";}return 0;}