[BFS][链表][二分][STL]JZOJ 5875 听我说，海蜗牛

Description

 

Input

Output

 

Sample Input

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

 

Sample Output

2
3
2
 

 

Data Constraint

分析

这题我们可以用一个链表来搞未加入连通块的点，然后在一个vector里面二分找当前点是否被断边（其实最好用哈希）

然后就裸搜了

听从大佬王的建议用两条队列写了链表（我傻了？）

 

#pragma GCC optimize(2)
#include <iostream>
#include <cstdio>
#include <vector> 
#include <queue>
#include <algorithm>
using namespace std;
const int N=2e5+10;
struct Edge {
    int u,v;
}e[2*N];
int cnt;
int n,m,q;
vector<int> a[N/2];
queue<int> qv;

bool Cmp(Edge a,Edge b) {
    return a.v<b.v;
}

int Find(int p,int x) {
    int l=0,r=a[p].size()-1;
    if (r<l) return 2147483647;
    while (l<=r) {
        int mid=l+r>>1;
        if (a[p][mid]==x) return a[p][mid];
        if (a[p][mid]>x) r=mid-1;
        else l=mid+1;
    }
    return a[p][l];
}

void Bfs(int v0) {
    queue<int> qu,q;
    while (!qu.empty()) qu.pop();
    while (!q.empty()) q.pop();
    qu.push(v0);
    while (!qu.empty()) {
        int u=qu.front();qu.pop();
        while (!qv.empty()) {
            int v=qv.front();qv.pop();
            if (Find(u,v)!=v) qu.push(v);
            else q.push(v);
        }
        while (!q.empty()) {
            int p=q.front();q.pop();
            qv.push(p);
        }
    }
}

int main() {
    freopen("connect.in","r",stdin);
    freopen("connect.out","w",stdout);
    scanf("%d%d%d",&n,&m,&q);
    for (int i=1;i<=m;i++) {
        cnt++;
        scanf("%d%d",&e[cnt].u,&e[cnt].v);
        e[++cnt].u=e[cnt-1].v;e[cnt].v=e[cnt-1].u;
    }
    sort(e+1,e+cnt+1,Cmp);
    for (int i=1;i<=cnt;i++) a[e[i].u].push_back(e[i].v);
    for (int i=1;i<=q;i++) {
        int p;
        scanf("%d",&p);
        while (!qv.empty()) qv.pop();
        for (int j=1;j<=p;j++) {
            int x;
            scanf("%d",&x);
            qv.push(x);
        }
        int ans=0;
        while (!qv.empty()) {
            int u=qv.front();qv.pop();
            Bfs(u);
            ans++;
        }
        printf("%d\n",ans);
    }
    fclose(stdin);fclose(stdout);
}