欧拉回路 HDU

Ant Trip

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2873    Accepted Submission(s): 1141

Problem Description

Ant Country consist of N towns.There are M roads connecting the towns.

Ant Tony,together with his friends,wants to go through every part of the country.

They intend to visit every road , and every road must be visited for exact one time.However,it may be a mission impossible for only one group of people.So they are trying to divide all the people into several groups,and each may start at different town.Now tony wants to know what is the least groups of ants that needs to form to achieve their goal.

 

Input

Input contains multiple cases.Test cases are separated by several blank lines. Each test case starts with two integer N(1<=N<=100000),M(0<=M<=200000),indicating that there are N towns and M roads in Ant Country.Followed by M lines,each line contains two integers a,b,(1<=a,b<=N) indicating that there is a road connecting town a and town b.No two roads will be the same,and there is no road connecting the same town.

 

Output

For each test case ,output the least groups that needs to form to achieve their goal.

 

Sample Input

3 31 22 31 34 21 23 4

 

Sample Output

12
Hint
New ~~~ Notice: if there are no road connecting one town ,tony may forget about the town.In sample 1,tony and his friends just form one group,they can start at either town 1,2,or 3.In sample 2,tony and his friends must form two group.
 

 

一个连通图不会有奇数个奇数度顶点，画一画图可以知道，假如一个连通图所有点度数为偶数，答案为1，否则就是奇数度顶点的一半，累加求和。还不是很理解，慢慢搞吧。

#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;vector<int> vec;int vist[200010], deg[200010], odd[200010], par[200010];int n, m;void init(){    vec.clear();    memset(vist, 0, sizeof(vist));    memset(deg, 0, sizeof(deg));    memset(odd, 0, sizeof(odd));    for(int i = 1; i <= n; i++) par[i] = i;}int Find(int x){    return x==par[x] ? x : par[x]=Find(par[x]);}void unite(int x, int y){    int fx = Find(x);    int fy = Find(y);    if(fx == fy) return ;    par[fx] = par[fy];}int main(){    while(~scanf("%d%d", &n, &m)) {        init();        for(int i = 0; i < m; i++) {            int a, b;            scanf("%d%d", &a, &b);            deg[a]++;            deg[b]++;            unite(a, b);        }        for(int i = 1; i <= n; i++) {            int fi = Find(i);            if(!vist[fi]) {                vec.push_back(fi);                vist[fi] = 1;            }            if(deg[i]&1) odd[fi]++; // 有度数为奇数的则父节点+1        }        int ans = 0;        for(int i = 0; i < vec.size(); i++) {            int t = vec[i];            if(deg[t] == 0) continue;            if(odd[t] == 0) ans++;            else ans += odd[t]/2;        }        printf("%d\n", ans);    }}