D、City

最新推荐文章于 2021-08-13 20:50:34 发布

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本文介绍了一个经典的并查集应用案例，通过并查集数据结构解决随着边的增加，图中连通块数量的变化问题。具体实现包括并查集的查找与合并操作，并通过逆向思维从最后0条边开始逐步增加边来统计每次更新后的连通块数量。

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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

题目大意：有n个点，m条线，每条线连接两个点，由线连接起来的点若干个点算一块，问每删去一条线后，点被分成多少块。

必备知识：并查集：参考博客：并查集详解 (转)

可以看成从最后0条线开始，每增加一条线后，点被分成几块

#include <stdio.h>
#include <string.h>

int pre[10005];
int line[100005][2];
int ans[100005];

int find(int x)
{
    int r = x;
    while (pre[r] != r)
        r = pre[r];

    int j = x, k;

    while (pre[j] != r)
    {
        k = pre[j];
        pre[j] = r;
        j = k;
    }

    return r;
}

int join(int x,int y)
{

    int fx = find(x),fy = find(y);
    if (fx != fy)
    {
        pre[fx] = fy;
        return 1;
    }
    return 0;
}

int main()
{
    int n,m;
    while (scanf("%d%d",&n,&m) != EOF)
    {
        memset(pre,0,sizeof(pre));
        memset(line,0,sizeof(line));
        memset(ans,0,sizeof(ans));
        for (int i = 0; i < 10005; i++)
        {
            pre[i] = i;
        }

        for (int i = 0; i < m; i++)
        {
            scanf("%d%d",&line[i][0],&line[i][1]);
        }

        int sum = n;

        for (int i = 0; i < m; i++)
        {
            ans[i] = sum;
            sum -= join(line[m-i-1][0],line[m-i-1][1]);
        }

        for (int i = m-1; i >= 0; i--)
        {
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}