2015
07-17

# D-City

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.

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.

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.

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.


#include <iostream>
#include <stack>
#include <stdio.h>
using namespace std;
const int MAX_N = 10000 + 100;
const int MAX_M = 100000 + 100;
int p[MAX_N];
int _find(int x)
{
return p[x] == x ? x : (p[x] = _find(p[x]));
}
int n, m;
stack <int> s;
struct Edge
{
int u, v;
};
Edge edge[MAX_M];
int res[MAX_M];

int main()
{
while(scanf("%d%d", &n, &m) != EOF)
{
for(int i = 0; i < n; i++)
p[i] = i;
for(int i = 0; i < m; i++)
scanf("%d%d", &edge[i].u, &edge[i].v);
res[m - 1] = n;
for(int i = m - 1; i > 0; i--)
{
int a = _find(edge[i].u);
int b = _find(edge[i].v);
if(a != b)
{
p[a] = b;
res[i - 1] = res[i] - 1;
}
else
res[i - 1] = res[i];
}
for(int i = 0; i < m; i++)
printf("%d\n", res[i]);
}
return 0;
}