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D-City

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

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. 
 

/*
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 
并查集逆向推理,把所有的点连接成图
题意：给定一个图，不停的删除边，问每删除一条边后有多少个连通块
注意最后一次输出的一定为n,即顶点的个数，此时的每个点都是孤立的。
从后往前推理，对于输入的m对数据，如果开始不连通，则把他们连通起来，对应的连通块的数量也要减1
*/
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
using namespace std;
int par[10100];
struct data
{
    int x;
    int y;
} a[100010];
int n,m;
int ans[100010];
void Init()
{
    memset(ans,0,sizeof(ans));
    for(int i = 0; i <= n; i ++)
        par[i] = i;
    for(int i = 1; i <= m; i ++)
        scanf("%d%d",&a[i].x,&a[i].y);
}
int find(int x)
{
    if(par[x] == x)
        return x;
    else
        return par[x] = find(par[x]);
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        Init();
        ans[m] = n;
        ///逆向考虑，所以此处在建图
        ///ans数组保存上一步的连通块的个数
        for(int i = m; i >= 1; -- i)
        {
            int x = find(a[i].x);
            int y = find(a[i].y);
            ///说明此时不在一个连通块上面，连接，则上一步的连通块数量减少1
            if(x != y)
            {
                if(x < y)
                    par[x] = y;
                else
                    par[y] = x;
                ans[i-1] = ans[i] - 1;
            }
            else
                ans[i-1] = ans[i];
        }
        for(int i = 1; i <= m; i ++)
            printf("%d\n",ans[i]);
    }
    return 0;
}

HDU4496 D-City并查集逆向推理

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原文地址：http://blog.csdn.net/wuxiushu/article/details/51328901