Day4 - K - Ant Trip HDU - 3018

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. 

InputInput 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.OutputFor each test case ,output the least groups that needs to form to achieve their goal.Sample Input

3 3
1 2
2 3
1 3

4 2
1 2
3 4

Sample Output

1
2

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.

简述：一个有回路与通路的混合图，问几笔能画完，忽略孤立点。

思路：不打印解，就使用并查集找到每条路的"root"，无向图，统计每条路上奇度数的点，队伍数 = 回路数 + 通路数(奇数点/2)

代码如下：

const int maxm = ;

int degree[maxm], fa[maxm], vis[maxm], num[maxm], N, M;
vector<int> root;

void init() {
    memset(degree, , sizeof(degree)), memset(vis, , sizeof(vis)), memset(num, , sizeof(num));
    root.clear();
    for (int i = ; i <= N; ++i)
        fa[i] = i;
}

int Find(int x) {
    if(x == fa[x])
        return x;
    return fa[x] = Find(fa[x]);
}

void Union(int x,int y) {
    int fx = Find(x), fy = Find(y);
    if(fx != fy)
        fa[fy] = fx;
}

int main() {
    while(scanf("%d%d",&N,&M) != EOF) {
        init();
        for (int i = ; i < M; ++i) {
            int t1, t2;
            scanf("%d%d", &t1, &t2);
            degree[t1]++, degree[t2]++;
            Union(t1, t2);
        }
        for(int i = ; i <= N; ++i) {
            int f = Find(i);
            if(!vis[f]) {
                root.push_back(f);
                vis[f] = ;
            }
            if(degree[i] % ) {
                num[f]++;
            }
        }
        int sum = ;
        for (auto i = root.begin(); i != root.end(); ++i) {
            if(degree[*i] == )
                continue;
            else if (num[*i] == )
                sum++;
            else if (num[*i])
                sum += num[*i] / ;
        }
        printf("%d\n", sum);
    }
    return ;
}

为什么通路数是奇数点/2呢？(没学离散，后面填坑)直接画图就能证明必要性，回路也如此。