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Zoj1628--Diamond(Dfs《暴力》)
阅读量:5743 次
发布时间:2019-06-18

本文共 3146 字,大约阅读时间需要 10 分钟。

Diamond

Time Limit: 2 Seconds     
Memory Limit: 65536 KB

Diamond mine is a mini-game which is played on an 8 * 8 board as you can see below.

The board is filled with different colors of diamonds. The player can make one move at a time. A move is legal if it swaps two adjacent diamonds(not diagonally) and after that, there are three or more adjacent diamonds in a row or column with the same color. Those diamonds will be taken away and new diamonds will be put in their positions. The game continues until no legal moves exist.

Given the board description. You are going to determine whether a move is legal.

Input

The input contains several cases. Each case has exactly 9 lines. The first 8 lines each contains a string of 8 characters. The characters are 'R'(Red), 'O'(Orange), 'G'(Green), 'P'(Purple), 'W'(White), 'Y'(Yellow) or 'B'(Blue). All characters are uppercase. No 3 diamonds of the same color are initially in adjacent positions in a row or column. The last line has 4 integers in the form " row1 column1 row2 column2" describing the postions of the 2 diamonds that the player wants to swap. Rows are marked 1 to 8 increasingly from top to bottom while columns from left to right. Input is terminated by EOF.

Output

For each case, output "Ok!" if the move is legal or "Illegal move!" if it is not.

Sample Input

PBPOWBGW

RRPRYWWP
YGBYYGPP
OWYGGRWB
GBBGBGGR
GBWPPORG
PPGORWOG
WYWGYWBY
4 3 3 3
PBPOWBGW
RRPRYWWP
YGBYYGPP
OWYGGRWB
GBBGBGGR
GBWPPORG
PPGORWOG
WYWGYWBY
5 5 6 5

Sample Output

Ok!

Illegal move!

 


Author: PAN, Minghao

Source: ZOJ Monthly, June 2003

话说一不小心WA了一晚上, 我nm。

ac码:

#include 
#include
#include
using namespace std;char map[10][10];bool Dfs(int a, int b) //相当于往四个方向搜。{ int flag1 = 0, flag2 = 0; for(int i = a; i >= 1; i--) { if(map[i][b] == map[a][b]) flag1++; else break; } for(int i = a+1; i <= 8; i++) { if(map[i][b] == map[a][b]) flag1++; else break; } if(flag1 >= 3) return true; for(int i = b; i >= 1; i--) { if(map[a][i]==map[a][b]) flag2++; else break; } for(int i = b + 1; i <= 8; i++) { if(map[a][i]== map[a][b]) flag2++; else break; } if(flag2 >= 3) return true; return false;}int main(){ while(cin >> map[1][1]) { for(int i = 1; i <= 8; i++) for(int j = 1; j <= 8; j++) { if(i == 1 && j == 1) continue; else cin >> map[i][j]; } int a, b, c, d; scanf("%d %d %d %d", &a, &b, &c, &d); if(abs(a-c) + abs(b-d) != 1){ //xiaojianzhi printf("Illegal move!\n"); continue; } char s; s = map[a][b];map[a][b] = map[c][d]; map[c][d] = s; // swap position. if(Dfs(a, b) || Dfs(c, d)) printf("Ok!\n"); else printf("Illegal move!\n"); } return 0; }

 

转载于:https://www.cnblogs.com/soTired/p/4759033.html

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