hdu 2120 Ice_cream's world I

2021-07-29 01:45:12

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=2120

Ice_cream's world I

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

8 10
0 1
1 2
1 3
2 4
3 4
0 5
5 6
6 7
3 6
4 7

Sample Output

用并查集统计环的个数。。

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstring>

#include<cstdio>

#include<vector>

#include<map>

using std::map;

using std::min;

using std::find;

using std::pair;

using std::vector;

using std::multimap;

#define pb(e) push_back(e)

#define sz(c) (int)(c).size()

#define mp(a, b) make_pair(a, b)

#define all(c) (c).begin(), (c).end()

#define iter(c) __typeof((c).begin())

#define cls(arr, val) memset(arr, val, sizeof(arr))

#define cpresent(c, e) (find(all(c), (e)) != (c).end())

#define rep(i, n) for(int i = 0; i < (int)n; i++)

#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)

const int N = 1010;

const int INF = 0x3f3f3f3f;

struct UinonFind {

    int ans, par[N], rank[N];

    inline void init(int n) {

        ans = 0;

        rep(i, n + 1) {

            par[i] = i;

            rank[i] = 0;

        }

    }

    inline int find(int x) {

        while(x != par[x]) {

            x = par[x] = par[par[x]];

        }

        return x;

    }

    inline void unite(int x, int y) {

        x = find(x), y = find(y);

        if(x == y) { ans++; return; }

        if(rank[x] < rank[y]) {

            par[x] = y;

        } else {

            par[y] = x;

            rank[x] += rank[x] == rank[y];

        }

    }

    inline void solve(int n, int m) {

        init(n);

        int x, y;

        while(m--) {

            scanf("%d %d", &x, &y);

            unite(x, y);

        }

        printf("%d\n", ans);

    }

}go;

int main() {

#ifdef LOCAL

    freopen("in.txt", "r", stdin);

    freopen("out.txt", "w+", stdout);

#endif

    int n, m;

    while(~scanf("%d %d", &n, &m)) {

        go.solve(n, m);

    }

    return 0;

}

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