DZY loves chemistry, and he enjoys mixing chemicals.
DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order.
Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
The first line contains two space-separated integers n and m .
Each of the next m lines contains two space-separated integers xi and yi (1 ≤ xi < yi ≤ n). These integers mean that the chemical xi will react with the chemical yi. Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 to n in some order.
Print a single integer — the maximum possible danger.
1 0
1
2 1 1 2
2
3 2 1 2 2 3
4
In the first sample, there's only one way to pour, and the danger won't increase.
In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.
In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).
题意:有n种药物,其中m对能发生化学反应,容器初始危险度为1,当放入一种药品若和容器中已有物品能发生反应,危险度就乘2,求按一定顺序放入所有药品危险度最大是多少。
思路:能发生反应的药品连成一棵树,同一棵树中的药品按照一定顺序加入就一定可以保证每次加入都发生反应。假设能分成3棵树,数量分别为a,b,c,各个反应次数为a-1,b-1,c-1,又a+b+c=n,所以总反应次数为n-树数。
# include <stdio.h> # include <math.h> int pre[51]; int find(int x) { if(x != pre[x]) pre[x] = find(pre[x]); return pre[x]; } int main() { int n, m, a, b; while(~scanf("%d%d",&n,&m)) { int road = 0; for(int i=0; i<=n; ++i) pre[i] = i; while(m--) { scanf("%d%d",&a,&b); int px = find(a); int py = find(b); if(px != py) { ++road; pre[px] = py; } } printf("%.0f\n",pow(2,road)); } return 0; }