UVa 705 - Slash Maze

題目：有一個二維的迷宮，墻和格子都是斜著放置的，求裡面最大的獨立區域面積。

分析：圖論、搜索。

            

            如上圖所示，這裡將一個斜著的正方形，分成兩個三角形部分，每個讀入字符的區域分成三個三角形；

            這樣可以構成4*row*column個小三角形，每個三角形和三個三角形相鄰；

            定義麼個節點有三個link指針，指向相鄰的三個三角形，如果是邊界或者有墻標記成-1；

            計算時，先把周圍的掃描一遍，清理掉所有的非封閉區域；

            然後，直接dfs即可；

            有些做法，將邊看成是有厚度的用2*2的矩陣表示一個圖形搜索；

說明：感覺寫的有點長。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;string wall[80];typedef struct _bnode{int link[3];}block;block Block[100001];bool  Visit[100001];int dfs(int v){if (v == -1 || true == Visit[v])return 0;Visit[v] = true;int ans = 1;for (int i = 0; i < 3; ++ i)ans += dfs(Block[v].link[i]);return ans;}int main(){int row, column, cases = 1;while (cin >> column >> row && row) {getchar();for (int i = 0; i < row; ++ i)getline(cin, wall[i]);//create graphfor (int i = 0; i < row; ++ i) {for (int j = 0; j < column; ++ j) {int u = (i*column+j)*4;int l = u+1, r = u+2, d = u+3;if (i != 0)Block[u].link[0] = d-column*4;else Block[u].link[0] = -1;if (wall[i][j] != '/')Block[u].link[1] = r;else Block[u].link[1] = -1;if (wall[i][j] != '\\')Block[u].link[2] = l;else Block[u].link[2] = -1;if (j != 0)Block[l].link[0] = r-4;else Block[l].link[0] = -1;if (wall[i][j] != '/')Block[l].link[1] = d;else Block[l].link[1] = -1;if (wall[i][j] != '\\')Block[l].link[2] = u;else Block[l].link[2] = -1;if (j != column-1)Block[r].link[0] = l+4;else Block[r].link[0] = -1;if (wall[i][j] != '/')Block[r].link[1] = u;else Block[r].link[1] = -1;if (wall[i][j] != '\\')Block[r].link[2] = d;else Block[r].link[2] = -1;if (i != row-1)Block[d].link[0] = u+column*4;else Block[d].link[0] = -1;if (wall[i][j] != '/')Block[d].link[1] = l;else Block[d].link[1] = -1;if (wall[i][j] != '\\')Block[d].link[2] = r;else Block[d].link[2] = -1;}}//non enclosed areamemset(Visit, 0, sizeof(Visit));for (int i = 0; i < row; ++ i) {for (int j = 0; j < column; ++ j) {int u = (i*column+j)*4;int l = u+1, r = u+2, d = u+3;if (i == 0)dfs(u);if (j == 0)dfs(l);if (j == column-1)dfs(r);if (i == row-1)dfs(d);}}int max = 0, ans, count = 0;for (int i = 0; i < row; ++ i) {for (int j = 0; j < column; ++ j) {int u = (i*column+j)*4;int l = u+1, r = u+2, d = u+3;if (i != 0 && !Visit[u]) {ans = dfs(u);if (max < ans)max = ans;count += ans>0;}if (j != 0 && !Visit[l]) {ans = dfs(l);if (max < ans)max = ans;count += ans>0;}if (j != column-1 && !Visit[r]) {ans = dfs(r);if (max < ans)max = ans;count += ans>0;}if (i != row-1 && !Visit[d]) {ans = dfs(d);if (max < ans)max = ans;count += ans>0;}}}printf("Maze #%d:\n",cases ++);if (count > 0)printf("%d Cycles; the longest has length %d.\n\n",count,max/2);else printf("There are no cycles.\n\n");}     return 0;}