演算法筆記之 並查集入門 POJ 1611

http://poj.org/problem?id=1611

Description

Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others. 
In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects
the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP). 
Once a member in a group is a suspect, all members in the group are suspects. 
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.

Input

The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student is numbered by a unique integer
between 0 and n−1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group.
Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space. 
A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.

Output

For each case, output the number of suspects in one line.

Sample Input

100 42 1 25 10 13 11 12 142 0 12 99 2200 21 55 1 2 3 4 51 00 0

Sample Output

411

並查集的介紹大家還是google吧，我只會貼代碼，見諒啊~

#include <iostream>using namespace std;int father[30001]; //儲存father indexint group[30001]; //儲存當前  集合的人數//初始化void init(int len) {for (int i = 0; i <= len; i++){father[i] = i;group[i] = 1;}}//尋找i的最終fatherint find_set(int i){if(i != father[i]){father[i] = find_set(father[i]); //路徑壓縮}return father[i];}//合并兩個點void join(int x,int y){int i = find_set(x);int j = find_set(y);if(i != j){father[i] = father[j];group[j] += group[i]; //合并時，由於最終father儲存當前集合人數}}int main() {int n,m,k;while( cin >> n >> m){if(n == 0)break;init(n); //初始化for(int i=0; i<m; i++){cin >> k;int pre,cur; //前一個 和 當前for(int j=0; j<k; j++){cin >> cur;if(j) //第一個數不用管join(pre,cur);pre =cur;}}//測試//for(int i=0; i<n; i++)//cout << father[i] << " ";//cout << endl;////for(int i=0; i<n; i++)//cout << group[i] << " ";//cout << endl;////cout << find_set(0) << endl;cout << group[find_set(0)] << endl;}return 0;}

已耗用時間：47ms 太慢了。 最佳化一下 遞迴和輸入輸出：

#include <stdio.h>int father[30001]; //儲存father indexint group[30001]; //儲存當前  集合的人數//初始化void init(int len) {for (int i = 0; i <= len; i++) {father[i] = i;group[i] = 1;}}//尋找i的最終fatherint find_set(int i) {int temp = i;while (temp != father[temp]) {temp = father[temp];}father[i] = temp;return temp;//if(i != father[i]){//father[i] = find_set(father[i]); //路徑壓縮//}//return father[i];}//合并兩個點void join(int x, int y) {int i = find_set(x);int j = find_set(y);if (i != j) {father[i] = father[j];group[j] += group[i]; //合并時，由於最終father儲存當前集合人數}}int main() {int n, m, k;while (scanf("%d %d", &n, &m) != EOF) {if (n == 0)break;init(n); //初始化for (int i = 0; i < m; i++) {scanf("%d", &k);int pre, cur; //前一個 和 當前for (int j = 0; j < k; j++) {scanf("%d", &cur);if (j) //第一個數不用管join(pre, cur);pre = cur;}}//測試//for(int i=0; i<n; i++)//cout << father[i] << " ";//cout << endl;////for(int i=0; i<n; i++)//cout << group[i] << " ";//cout << endl;////cout << find_set(0) << endl;printf("%d\n", group[find_set(0)]);}return 0;}