Hdu 2120 Ice_cream ' s World I

Topic Connection

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

Ice_cream ' s World idescription

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 is some watchtowers is set up, and wall between watchtowers is build, in order to partition the Ice_cream ' s WOR Ld. But what many acmers at most can be awarded by the Queen is a big problem. One wall-surrounded land must is given to only one acmer and no walls is crossed if you can help the Queen solve this PR Oblem, you'll be get-a land.

Input

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

Output

Output the maximum number of acmers who'll 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

3

Use and check the number of statistical rings.

#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::p air; 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, si Zeof (arr)) #define Cpresent (C, E) (Find (All (c), (e))! = (c). End ()) #define REP (i, n) for (int i = 0; i < (int) n; i++) #defi NE tr (c, I) for (ITER (c) i = (c). Begin (); I! = (c). end (); ++i) const int N = 1010;const int INF = 0x3f3f3f3f;struct Uinonfin    d {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; } inlinevoid 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;}

Hdu 2120 Ice_cream ' s World I