题意翻译
DZY热爱化学.他有n种物质,其中m对会反应.他把它们一种一种倒到烧杯里.有一个危险值,一开始等于1.如果一种物质倒到烧杯里之后至少有一种物质能和它反应,则危险值乘以二;否则危险值不变.求所有物质倒到烧杯里之后最大的危险值.
By @Fuko_Ibuki
题目描述
DZY loves chemistry, and he enjoys mixing chemicals.
DZY has n n n chemicals, and m m 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 1 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 2 2 . Otherwise the danger remains as it is.
Find the maximum possible danger after pouring all the chemicals one by one in optimal order.
输入输出格式
输入格式：
The first line contains two space-separated integers n n n and m m m 
.
Each of the next m m m lines contains two space-separated integers xi x_{i} xi and yi y_{i} yi (1<=xi<yi<=n) (1<=x_{i}<y_{i}<=n) (1<=xi<yi<=n) . These integers mean that the chemical xi x_{i} xi will react with the chemical yi y_{i} yi . Each pair of chemicals will appear at most once in the input.
Consider all the chemicals numbered from 1 1 1 to n n n in some order.
输出格式：
Print a single integer — the maximum possible danger.
输入输出样例
输入样例#1： 复制
1 0
输出样例#1： 复制
1
输入样例#2： 复制
2 1 1 2
输出样例#2： 复制
2
输入样例#3： 复制
3 2 1 2 2 3
输出样例#3： 复制
4
说明
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 1 1 1 st chemical first, or pour the 2 2 2 nd chemical first, the answer is always 2 2 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).
用并查集维护即可；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 100005
#define inf 0x3f3f3f3f
#define INF 999999999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
int n, m;
ll ans;
int fa[maxn];
void init() {
for (int i = 0; i <= n; i++)fa[i] = i;
}
int findfa(int x) {
if(x == fa[x])return x;
return fa[x] = findfa(fa[x]);
}
void merge(int x, int y) {
if (findfa(x) != findfa(y)) {
fa[findfa(x)] = findfa(y);
}
}
int main()
{
//ios::sync_with_stdio(false);
rdint(n); rdint(m);
init(); ll ans = 1;
while (m--) {
int x, y;
rdint(x); rdint(y);
merge(x, y);
}
int tot = 0;
for (int i = 1; i <= n; i++) {
if (fa[i] == i)tot++;
}
cout << ((ll)1 << (n - tot)) << endl;
return 0;
}