首页 > ACM题库 > HDU-杭电 > HDU 4496-D-City-并查集-[解题报告]HOJ
2015
07-17

HDU 4496-D-City-并查集-[解题报告]HOJ

D-City

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

5 10 
0 1 
1 2 
1 3 
1 4 
0 2 
2 3 
0 4 
0 3 
3 4 
2 4

样例输出:

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.

题目链接:HDU 4496 D-City

并查集。

先读入边,再倒着添加边就行了。

#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;
}

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参考:http://blog.csdn.net/fobdddf/article/details/23303263