HDU Problem 2120 Ice_cream's world I 【并查集检查环】

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1193 Accepted Submission(s): 716

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.

Output

Output the maximum number of ACMers who will be awarded.
One answer one line.

Sample Input

Sample Output

3

Author

Wiskey

Source

HDU 2007-10 Programming Contest_WarmUp

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题意：有m个围墙连接n个哨岗，求由围墙包围得有多少个环。

思路：利用并查集，当两个哨岗的父节点相同时必定成一个环，检查共有多少个环。

#include <bitsstdc++.h>
#define MAX_N 1005
using namespace std;
const int INF = 1e9;

int par[MAX_N], m, n, ans;
void init() {
    for (int i = 0; i < 1002; i++) {
        par[i] = i;
    }
}
int Find(int x) {
    if (x == par[x]) return x;
    return par[x] = Find(par[x]);
}
void unite(int x, int y) {
    int fx = Find(x);
    int fy = Find(y);
    if (fx == fy) ans++;
    else par[fy] = fx;
}
int main() {
    int a, b;
    while (scanf("%d%d", &n, &m) != EOF) {
        init(); ans = 0;
        for (int i = 0; i < m; i++) {
            scanf("%d%d", &a, &b);
            unite(a, b);
        }
        printf("%d
", ans);
    }
    return 0;
}