并查集：HDU4496-D-City（倒用并查集）

D-City

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2448 Accepted Submission(s): 862

Problem Description

Luxer is a really bad guy. He destroys everything he met.
One day Luxer went to D-city. D-city has N D-points and M D-lines. Each D-line connects exactly two D-points. Luxer will destroy all the D-lines. The mayor of D-city wants to know how many connected blocks of D-city left after Luxer destroying the first K D-lines in the input.
Two points are in the same connected blocks if and only if they connect to each other directly or indirectly.

Input

First line of the input contains two integers N and M.
Then following M lines each containing 2 space-separated integers u and v, which denotes an D-line.
Constraints:
0 < N <= 10000
0 < M <= 100000
0 <= u, v < N.

Output

Output M lines, the ith line is the answer after deleting the first i edges in the input.

Sample Input

5 10
0 1
1 2
1 3
1 4
0 2
2 3
0 4
0 3
3 4
2 4

Sample Output

1
1
1
2
2
2
2
3
4
5

Hint

The graph given in sample input is a complete graph, that each pair of vertex has an edge connecting them, so there’s only 1 connected block at first.
The first 3 lines of output are 1s because after deleting the first 3 edges of the graph, all vertexes still connected together.
But after deleting the first 4 edges of the graph, vertex 1 will be disconnected with other vertex, and it became an independent connected block.
Continue deleting edges the disconnected blocks increased and finally it will became the number of vertex, so the last output should always be N.

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解题心得：

-题意就是给你一个图，m个边，从第一个边开始每次取消一个，问你每次取消之后有多少个单独的集合
-这个题看起来挺唬人的，其实就是将给出的点拿来从从后面开始合并就可以了。只要是不同源的相合并，那么就必然少一个单独的集合，思维不要僵化，换个方向想就可以了。

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
const int maxn2 = 1e4+10;
int n,m,father[maxn2];
stack <int> st;//因为是从后面开始合并，所以用栈来存放
struct node
{
    int a,b;
}edge[maxn];

int find(int a)
{
    if(father[a] == a)
        return a;
    else
        return father[a] = find(father[a]);
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<=n;i++)
            father[i] = i;
        for(int i=0;i<m;i++)
            scanf("%d%d",&edge[i].a,&edge[i].b);

        int ans = n;
        st.push(n);
        for(int i=m-1;i>0;i--)//从后往前开始合并，注意第一条边并没有实际存在，因为在一开始就已经被破坏了
        {
            int fa = find(edge[i].a);
            int fb = find(edge[i].b);
            if(fa != fb)//不同源，集合必然少一个
            {
                father[fa] = fb;
                ans--;
            }
            st.push(ans);
        }

        while(!st.empty())
        {
            printf("%d
",st.top());
            st.pop();
        }
    }
    return 0;
}