标签：GDUT 21 int 排位赛 number four ++ board test

题意

Feras bought to his nephew Saleem a new game to help him learning calculating. The game consists of a board with 4 rows and 4 columns with 16 cubes. Every cube has a number from 1 to 16. Let's define the power of a column as the sum of its elements. In the same way, the power of a row is the sum of its elements. Saleem should arrange the cubes in the board such that the power of all columns and all rows are equal. To make the game easier, the nice uncle, Feras, will help him arranging 7 cubes, and Saleem should arrange the rest of the cubes.

输入格式

Your program will be tested on one or more test cases. The first line of the input will be a single integer T, the number of test cases (1 ≤  T ≤  100). Then the test cases. Each test case has four lines containing four integers. The j-th number in the i-th line describes the cell (i,j) of the board. If the number is -1 then the cell is empty and you have to fill it, otherwise, uncle Feras has already filled this cell.

输出格式

For each test case print a line in the following format: "Case c:" where c is the test case number starting from 1 then print the board in four lines every line has four numbers separated by space. If there is more than one solution print the solution that has the smallest order (See the notes below).

样例1

思路

深搜。

每次搜到填完的时候判断一下符不符合条件，具体做法也很简单，就是开个数组t[8]分别记录4行和4列的数的和是否相等。

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 9;
int a[N][N];
int vis[20];

bool is_ok() {
    int t[8] = {0};
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 4; j++) {
            t[i] += a[i][j];
            t[j+4] += a[i][j];
        }
    }
    for (int i = 1; i < 8; i++) {
        if (t[i] != t[i-1]) return 0;
    }
    return 1;
}

bool dfs(int step) {
    if (step == 16) {
        if (is_ok()) return 1;
        return 0;
    } else {
        if (a[step/4][step%4] == -1) {
            for (int i = 1; i <= 16; i++) {
                if (!vis[i]) {
                    vis[i] = 1;
                    a[step/4][step%4] = i;
                    if (dfs(step+1)) return 1;
                    vis[i] = 0;
                    a[step/4][step%4] = -1;
                }
            }
        } else if (dfs(step+1)) return 1;
        return 0;
    }
}

int main() {
    int t;
    cin >> t;
    int kase = 1;
    while (t--) {
        memset(vis,0,sizeof vis);
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 4; j++) {
                cin >> a[i][j];
                if (a[i][j] != -1) vis[a[i][j]] = 1;
            }
        }
        dfs(0);
        printf("Case %d:\n", kase++);
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 4; j++) {
                if (j == 0) printf("%d", a[i][j]);
                else printf(" %d", a[i][j]);
            }
            puts("");
        }
    }
}

标签：GDUT,21,int,排位赛,number,four,++,board,test
来源： https://www.cnblogs.com/zenghaifan/p/16068161.html

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