hdu4496 D-City 並查集

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，原來有t個連通分支，然後不斷地往裏面加邊，如果邊的兩個端點原來不在同一個連通分支裏面，那麼把它們兩個的連通分支合爲一個，且t = t - 1，如果原來就在同一個連通分支裏面，對t不操作，用並查集來判斷是否在同一個連通分支裏面以及合併操作，用一個數組記錄t的值用於最後輸出，注意逆向加邊的時候，最後會多個t = 1，最後要去掉

#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;
}