P3144 关闭农场 并查集 反向

  

FJ和他的奶牛们正在计划离开小镇做一次长的旅行,同时FJ想临时地关掉他的农场以节省一些金钱。

这个农场一共有被用M条双向道路连接的N个谷仓(1<=N,M<=3000)。为了关闭整个农场,FJ 计划每一次关闭掉一个谷仓。当一个谷仓被关闭了,所有的连接到这个谷仓的道路都会被关闭,而且再也不能够被使用。

FJ现在正感兴趣于知道在每一个时间(这里的“时间”指在每一次关闭谷仓之前的时间)时他的农场是否是“全连通的”——也就是说从任意的一个开着的谷仓开始,能够到达另外的一个谷仓。注意自从某一个时间之后,可能整个农场都开始不会是“全连通的”。

输入输出样例

输入样例#1: 复制

4 3
1 2
2 3
3 4
3
4
1
2

输出样例#1: 复制

YES
NO
YES
YES

题意:给出n个点和m条无向边 每次去掉一个点查看联通块是否为1

遵循先判断输出 然后去掉一个点的原则

并查集find1一定要优化  一开始就因为这个超时

因为并查集没有拆掉的操作  所以要倒着来  每次加上一个点  (比如拆到最后一个点的时候是一个点   倒着来补上第一个点的时候也是那一个点   所以是一样的)
因为奇特的读入输出模式  其实就是为了倒着来所准备的

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);i--)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define LL long long
#define pb push_back
#define fi first
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
///////////////////////////////////
#define inf 0x3f3f3f3f
#define N 3000+5

int n,m;
int x[N],y[N];
int in[N];
int f[N];
int node[N];
int ans[N];
int find1(int x)
{
    return f[x]==x?x:(f[x]=find1(f[x]) );

}
void union1(int a,int b)
{
    int x=find1(a);
    int y=find1(b);
    if(x!=y)
        f[x]=y;
}

int main()
{
    RII(n,m);
    rep(i,1,n)
    f[i]=i;
    rep(i,1,m)
    {
        RII(x[i],y[i]);
    }
    rep(i,1,n)
    {
        RI(node[i]);
        in[node[i]]=0;
    }

    repp(i,n,1)
    {
        in[ node[i] ]=1;
        rep(j,1,m)
        if( in[ x[j] ]&&in[ y[j] ] )
            union1( x[j],y[j] );

        rep(j,1,n)//遍历n查看有几个联通块
        if(in[j]&&f[j]==j)
            ans[i]++;
    }
    rep(i,1,n)
    if(ans[i]==1)
        printf("YES\n");
    else
        printf("NO\n");
   return 0;
}

原文地址:https://www.cnblogs.com/bxd123/p/10569118.html

时间: 2024-11-08 18:27:46

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