D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2649    Accepted Submission(s): 922

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.

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

Recommend

liuyiding   |   We have carefully selected several similar problems for you:  5498 5497 5496 5495 5494

求联通块数量 ，  反向思维， 很神奇。

#include <cstdio>
#include <cstring>
#include <iostream>
#define N 10001
#define M 100001
using  namespace std;
int n, m;
int a[M], b[M], num[M], father[N];
void Init()
{
    num[m-1] = n;
    for(int i = 0; i < n; i++)
        father[i] = i;
}
int Find(int a)
{
    if(a == father[a])
        return a;
    else
        return father[a] = Find(father[a]);
}
void Mercy(int a, int b, int c)
{
    int Q = Find(a);
    int P = Find(b);
    if(Q != P)
    {
        father[Q] = P;
        num[c-1] = num[c] - 1;
    }
    else
        num[c-1] = num[c];
}
int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        Init();
        for(int i = 0; i < m; i++)
            scanf("%d%d", &a[i], &b[i]);
        for(int i = m-1; i >= 0; i--)   //反向 ;
            Mercy(a[i], b[i], i);
        for(int i = 0; i < m; i++)
            printf("%d\n", num[i]);
    }
    return 0;
}