Codeforces445B_DZY Loves Chemistry(DFS/并查集)

DZY Loves Chemistry

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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 xiwill 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.

Output

Print a single integer — the maximum possible danger.

Sample test(s)

input

1 0

output

1

input

2 1
1 2

output

2

input

3 2
1 2
2 3

output

4

Note

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).

解题报告

意思非常不清楚。

。。

有一试管。有N个化学物品，M对能发生化学反应的药品。求最大危急系数。

空试管的危急系数为1，当有一个药品或多个药品发生反应的时候。危急系数乘以2。

那样的话对于任一联通块。都会产生危急×2

一開始以为找最大的联通块。。。

这题能够用并查集求出联通块。。

。

也能够直接DFS搜

#include <iostream>
#include <cstring>
#include <cstdio>
#define N 55
#define M 12550
using namespace std;
struct node
{
    int v,next;
} edge[M];
int head[N],n,m,cnt,maxx=0,vis[N];
void add(int u,int v)
{
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
void  dfs(int s)
{
    vis[s]=1;
    for(int i=head[s]; i!=-1; i=edge[i].next)
    {
        if(!vis[edge[i].v])
        {
            maxx++;
            dfs(edge[i].v);
        }
    }
}
int main()
{
    int i,j,u,v;
    while(~scanf("%d%d",&n,&m))
    {
        cnt=0;
        maxx=0;
        __int64 dd=1;
        memset(edge,0,sizeof(edge));
        memset(head,-1,sizeof(head));
        memset(vis,0,sizeof(vis));
        for(i=0; i<m; i++)
        {
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
        int ans=0;
        for(i=1; i<=n; i++)
        {
            dfs(i);
        }
        printf("%I64d
",dd<<maxx);
    }
    return 0;
}

并查集代码

#include <iostream>
#include <cstdio>
#include <cstring>
#define N 55
#define M 12500
using namespace std;
int B[N],num[N];
int fine(int x)
{
    if(B[x]!=x)
    B[x]=fine(B[x]);
    return B[x];
}
void bin(int x,int y)
{
    int xx=fine(x);
    int yy=fine(y);
    if(xx!=yy)
    {
        B[xx]=yy;
        num[xx]+=num[yy];
        num[yy]=1;
    }
}
int n,m;
int main()
{
    int i,j,u,v;
    while(~scanf("%d%d",&n,&m))
    {
        for(i=1;i<=n;i++)
        {
            B[i]=i;
            num[i]=1;
        }
        for(i=0;i<m;i++)
        {
            scanf("%d%d",&u,&v);
            bin(u,v);
        }
        int sum=0;
        for(i=1;i<=n;i++)
        {
            if(num[i]!=1)
            {
                sum+=num[i]-1;
            }
        }
        __int64 d=1;
        printf("%I64d
",d<<sum);
    }
    return 0;
}