(Water question NUPT 1593) 8. Queen's question (judge whether any element is in the same row, column, or diagonal line), nupt1593
Question:
8. Queen's questionTime limit (Common/Java):2000 MS/6000 MS running memory limit: 65536 KByte
Total submissions: 1116 pass the test: 137
Description
Put eight queens on 8x8 garbled chess. Please judge whether they attack each other, that is, whether both queens are in the same row, column, or diagonal line.
Input
The input contains multiple test cases (up to 104 ). The number of test cases is given first. Then there is each test case. Each test case contains eight lines, each line uses eight characters to indicate the placement of the Queen on the board, "Q" indicates the queen, and "#" indicates null.
Output
For each test case, the output line contains:
L "Case #:", # indicates the serial number
L if the eight queens in the test case attack each other, the output is Yes; otherwise, the output is No.
Sample Input
2
Q #######
### Q ####
##### Q ##
# Q #####
####### Q
#### Q ###
###### Q #
# Q #####
####### Q
### Q ####
Q #######
# Q #####
##### Q ##
# Q ######
###### Q #
#### Q ###
Sample output
Case 1: Yes
Case 2: No
Prompt
Question Source
NUPT
Question Analysis:
Simple question. Because the same diagonal line was missing at the beginning, it was later added to determine whether the line was the same. So it seems a little messy. However, the AC is okay.
The Code is as follows:
/** Aa2.cpp ** Created on: March 24, 2015 * Author: Administrator */# include <iostream> # include <cstdio> # include <cmath> using namespace std; const int maxn = 8; struct Point {int x; int y;} points [maxn]; char map [maxn + 1] [maxn + 1]; bool visited [maxn]; int main () {int t; scanf ("% d", & t); int kk; for (kk = 1; kk <= t; ++ kk) {// printf ("Case % d:", kk); memset (visited, false, sizeof (visited); int I; for (I = 0; I <maxn; ++ I) {scanf ("% s", & map [I]);} bool flag = true; // checks whether a specific element is in the same column of int j; for (I = 0; I <maxn; ++ I) {for (j = 0; j <maxn; ++ j) {if (map [I] [j] = 'q') {points [I]. x = I; points [I]. y = j; if (visited [j] = false) {visited [j] = true;} else {flag = false;} break ;}} if (flag = false) {break ;}// you can determine whether a specific element is on the same diagonal line. otherwise, WA) if (flag = true) {for (I = 0; I <maxn; ++ I) {for (j = I + 1; j <maxn; ++ j) {if (fabs (points [I]. x-points [j]. x) = fabs (points [I]. y-points [j]. y) {flag = false ;}}} if (flag = true) {printf ("Case % d: No \ n", kk );} else {printf ("Case % d: Yes \ n", kk) ;}} return 0 ;}