hdu4496 D-City 并查集

D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3933    Accepted Submission(s): 1417

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.

逆向看，原来有t个连通分支，然后不断地往里面加边，如果边的两个端点原来不在同一个连通分支里面，那么把它们两个的连通分支合为一个，且t = t - 1，如果原来就在同一个连通分支里面，对t不操作，用并查集来判断是否在同一个连通分支里面以及合并操作，用一个数组记录t的值用于最后输出，注意逆向加边的时候，最后会多个t = 1，最后要去掉

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <set>

using namespace std;

const int maxn = 10005, maxm = 100005;
int T, n, m;
int f[maxn], ans[maxm], len;

struct Edge {
	int u, v;
	Edge() {}
	Edge(int u, int v) : u(u), v(v) {}
}edges[maxm];

int getf(int x) {
	return x == f[x] ? x : f[x] = getf(f[x]);
}

void Merge(int x, int y) {
	int u = getf(x), v = getf(y);
	if (u != v) {
		if (u < v) {
			f[v] = u;
		}
		else {
			f[u] = v;
		}
	}
}

int main()
{
	while (~scanf("%d%d", &n, &m)) {
		for (int i = 0; i < n; i++) {
			f[i] = i;
		}
		for (int i = 0; i < m; i++) {
			scanf("%d%d", &edges[i].u, &edges[i].v);
		}
		int t = n;
		ans[0] = n;
		len = 1;
		for (int i = m - 1; i >= 0; i--) {
			int u, v, x, y;
			u = edges[i].u;
			v = edges[i].v;
			x = getf(u);
			y = getf(v);
			if (x != y) {
				t -= 1;
			}
			ans[len++] = t;
			Merge(u, v);
		}
		for (int i = len - 2; i >= 0; i--) {
			printf("%d\n", ans[i]);
		}
	}
	return 0;
}