uva 657

哎，本来以为是一道简单题，但是WA了5，6次。。。这道题说给一个骰子的图，其中‘.’表示背景，‘*’表示是骰子的一个面，‘X’表示骰子面中的点。其中面和骰子的点都是4连接的，也就是说考虑上下左右四个方向。思路应该比较清晰：首先对骰子的面进行搜索，碰到面中的‘X’再对‘X’进行搜索。注意：对‘X’进行搜索时，需要注意‘X’不仅和相邻的‘X’是连接的，‘X’和相邻‘*’也是连接的（属于骰子的同一个面）。在我的代码中将搜索过的‘X’转换为尚未访问的‘*’，之后再对该点进行搜索。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <memory.h>#include <string>#include <string.h>#include <algorithm>using namespace std;const int maxn = 50 + 20;char dice[maxn][maxn];int visitted[maxn][maxn];int diceNum[maxn], diceBlocks;int row, col, diretions[4][2] = { { -1, 0 }, { 0, -1 }, { 0, 1 }, {1, 0} };void dfsF(int r, int c);bool inArea(int r, int c){if (r < 0 || r > row || c < 0 || c > col)return false;return true;}bool check(int r, int c){if (dice[r][c] == '*' || dice[r][c] == 'X')return true;return false;}void dfsS(int r, int c){if (inArea(r, c) && !visitted[r][c] && dice[r][c] == 'X'){visitted[r][c] = false;dice[r][c] = '*';for (int i = 0; i < 4; i++)dfsS(r + diretions[i][0], c + diretions[i][1]);/*for (int i = 0; i < 4; i++)dfsF(r + diretions[i][0], c + diretions[i][1]);*/}}void dfsF(int r, int c){if (inArea(r, c) && !visitted[r][c] && check(r, c)){visitted[r][c] = true;if (dice[r][c] == 'X'){visitted[r][c] = false;dfsS(r, c);diceNum[diceBlocks]++;dfsF(r, c);}else{for (int i = 0; i < 4; i++)dfsF(r + diretions[i][0], c + diretions[i][1]);}}}int main(){freopen("in2.txt", "r", stdin);freopen("out.txt", "w", stdout);int kaseNum = 0;while (cin >> col >> row){memset(visitted, false, sizeof(visitted));memset(diceNum, 0, sizeof(diceNum));diceBlocks = 0;if (col <= 0 || row <= 0)break;for (int i = 0; i < row; i++)cin >> dice[i];for (int i = 0; i < row; i++){for (int j = 0; j < col; j++){if ('*' == dice[i][j] && !visitted[i][j]){dfsF(i, j);if (diceNum[diceBlocks] > 0 && diceNum[diceBlocks] <= 6)diceBlocks++;}}}//for int isort(diceNum, diceNum + diceBlocks);cout << "Throw " << ++kaseNum << endl;for (int i = 0; i < diceBlocks - 1; i++)cout << diceNum[i] << " ";cout << diceNum[diceBlocks - 1] << endl << endl;}//whilereturn 0;}