输入格式

The first line is an integer N, the number of cases, 1<=N<=10
N lines follow. Each line contains 3 integers(n r c), The "n" stands for a square matrix in size of n * n.
The "r" is the row of the element. The "c" is the column of the element.  1 <= r, c <= n <= 100 

输出格式

For each case, output the scan sequence number of the element.

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
int n, r, c;
const int maxn = 1e2 + 20;
int a[maxn][maxn];
int tonext[2][2] = {{1, -1}, {-1, 1}};
void dfs(int num, int x, int y, int now) {
    a[x][y] = num;
    if (x == n && y == n) return;
    if (now == 0) { //减y的
        if (y == 1 && x != n) {
            dfs(num + 1, x + 1, y, !now);
        } else if (x == n) {
            dfs(num + 1, x, y + 1, !now);
        } else {
            dfs(num + 1, x + tonext[now][0], y + tonext[now][1], now);
        }
    } else { //减x的
        if (x == 1 && y != n) {
            dfs(num + 1, x, y + 1, !now);
        } else if (y == n) {
            dfs(num + 1, x + 1, y, !now);
        } else {
            dfs(num + 1, x + tonext[now][0], y + tonext[now][1], now);
        }
    }
}
void work() {
    scanf("%d%d%d", &n, &r, &c);
    a[1][1] = 1;
    dfs(2, 1, 2, 0);
//    for (int i = 1; i <= n; ++i) {
//        for (int j = 1; j <= n; ++j) {
//            printf("%d ", a[i][j]);
//        }
//        printf("\n");
//    }
    printf("%d\n", a[r][c]);
}

int main() {
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}