HDU4496 D-City并查集逆向推理

D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3035    Accepted Submission(s): 1074

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. 
 

 

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

/*Sample Input5 10 0 1 1 2 1 3 1 4 0 2 2 3 0 4 0 3 3 4 2 4  Sample Output1 1 1 2 2 2 2 3 4 5 并查集逆向推理,把所有的点连接成图题意：给定一个图，不停的删除边，问每删除一条边后有多少个连通块注意最后一次输出的一定为n,即顶点的个数，此时的每个点都是孤立的。从后往前推理，对于输入的m对数据，如果开始不连通，则把他们连通起来，对应的连通块的数量也要减1*/#include <iostream>#include <algorithm>#include <stdio.h>#include <string.h>using namespace std;int par[10100];struct data{    int x;    int y;} a[100010];int n,m;int ans[100010];void Init(){    memset(ans,0,sizeof(ans));    for(int i = 0; i <= n; i ++)        par[i] = i;    for(int i = 1; i <= m; i ++)        scanf("%d%d",&a[i].x,&a[i].y);}int find(int x){    if(par[x] == x)        return x;    else        return par[x] = find(par[x]);}int main(){    while(~scanf("%d%d",&n,&m))    {        Init();        ans[m] = n;        ///逆向考虑，所以此处在建图        ///ans数组保存上一步的连通块的个数        for(int i = m; i >= 1; -- i)        {            int x = find(a[i].x);            int y = find(a[i].y);            ///说明此时不在一个连通块上面，连接，则上一步的连通块数量减少1            if(x != y)            {                if(x < y)                    par[x] = y;                else                    par[y] = x;                ans[i-1] = ans[i] - 1;            }            else                ans[i-1] = ans[i];        }        for(int i = 1; i <= m; i ++)            printf("%d\n",ans[i]);    }    return 0;}