並查集練習之 POJ（709） Oil Deposits

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=709

The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.InputThe input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.OutputFor each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.Sample Input1 1*3 5*@*@***@***@*@*1 8@@****@*5 5 ****@*@@*@*@**@@@@*@@@**@0 0Sample Output0122

統計油田個數。 由於8個方向。實際上可以在讀入資料的時候只對左上四個方向進行 合并。 因為後面的也會合并上面的。

這個也可用DFS深度優先遍曆做。個人覺得效率要低點。

#include <iostream>using namespace std;int father[10001]; //把二維轉化為1維，使用並查集char map[101][101]; //void init(int len) {for (int i = 0; i <= len; i++)father[i] = i;}int find_set(int i) {if (i != father[i]) {father[i] = find_set(father[i]);}return father[i];}void join(int x, int y) {int i = find_set(x);int j = find_set(y);if (i != j)father[i] = father[j];}//只用尋找左上四個方向就行了（後面的還為讀取到）int dir[4][2] = { { -1, 0 }, { -1, -1 }, { 0, -1 }, { 1, -1 } };int main() {int m, n;//m為列  n為行while (cin >> m >> n) {if (m == 0)break;init(m * n); //初始化並查集//cout << find_set(2) << endl;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {//scanf("%c",&map[i][j]);cin >> map[i][j];//cout << map[i][j];int tempj, tempi;//如果當前讀入為 @， 就和左上比較，可以加入一個集合if (map[i][j] == '@') {for (int k = 0; k < 4; k++) {tempj = j + dir[k][0];tempi = i + dir[k][1];if (tempi >= 0 && tempi < m && tempj >= 0&& tempj < n) {//可以加入一個集合if (map[tempi][tempj] == '@') {join(i*n + j, tempi * n + tempj);}}}} else {father[i*n + j] = -1; //後面要統計並查集的集合。 設為-1不統計沒有油田的位置}}}//尋找有幾個集合int cnt = 0;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {int index = i*n + j;if (map[i][j] == '@' && father[index] != -1 && find_set(index) == (index)){cnt ++;}}}cout << cnt << endl;}return 0;}