hdu4496 D-City（反向並查集）

標籤：並查集

轉載請註明出處：http://blog.csdn.net/u012860063

題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=4496

D-City

Problem DescriptionLuxer 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. InputFirst 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.
 OutputOutput 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 5HintThe 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.  

 
Source2013 ACM-ICPC吉林通化全國邀請賽——題目重現 

/*題目不是很難，只需要逆向思考（正向行不通的時候，我們不防換一條路試試，生活亦是如此）；我們可以逆向認為所有的點全是獨立的，因為正向的時候去掉其中某條邊的，獨立的點不一定會增多（去掉這條邊後還有其他邊間接的相連），所以當我們逆向思考的時候，只會在增加某一條邊時減少獨立的點（也就是聯通的點增多），這樣只會在他之後才會有可能有某條邊的操作是“無效”的（聯通的點不變）；*///並查集！#include <cstdio>#include <iostream>#include <algorithm>using namespace std;#define M 10047int father[M];int ans[M];//記錄獨立的點數struct node{int l,r;}P[100047];int find(int x){return x == father[x]?x:father[x]=find(father[x]);}int main(){int n, m, i, j, t;while(~scanf("%d%d",&n,&m)){t = n;for(i = 0; i <= n; i++){father[i] = i;}for(i = m-1; i >= 0; i--)//注意此處的i是從大到小的，把去掉邊的操作逆向儲存在結構體中{ //這樣就能達到上面說得逆向思考scanf("%d%d",&P[i].l,&P[i].r);}for(i = 0; i < m; i++)   //因為記錄的邊已經逆向，這裡就不再需要{int f1 = find(P[i].l);int f2 = find(P[i].r);if(f2 != f1){father[f1] = f2;t--;            //當有新增的邊使獨立的點減少的時候}ans[i] = t;}for(i = m - 2; i >= 0; i--){printf("%d\n",ans[i]); //逆向輸出}printf("%d\n",n);}return 0;}