2013 ACM 通化邀请赛D．D

2013 ACM 通化邀请赛D．D

点击打开链接

D．D-City
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


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.

反向其实就是一个并查集的合并操作，。听说正向BFS可破，但是没试过

#include<stdio.h>
#include<string.h>
int dele[100010][2];
int father[10010];
int mystack[100010];
int top;
int count;
int getfather(int i)
{if(father[i] == i)return i;elsereturn  father[i] = getfather(father[i]);
}
void merge(int a, int b)
{a = getfather(a);b = getfather(b);if(a == b)return;else {father[b] = a;count --;}
}
int main()
{int n, m;while(scanf("%d%d", &n, &m) != EOF){int i;memset(dele, 0, sizeof(dele));for(i = 0; i < m; i++)father[i] = i;for(i = 0; i < m ; i++)scanf("%d%d", &dele[i][0], &dele[i][1]);count = n;top = 0;for(i--; i >= 0; i --){mystack[top++] = count;merge(dele[i][0], dele[i][1]);}for(top --; top >= 0; top--){printf("%dn", mystack[top]);}}return 0;
}