Leaving Auction CodeForces

大意: 若干个人参加拍卖会, 给定每个人出价顺序, 保证价格递增, q个询问, 给出k个人的编号, 求删除这k个人的所有出价后, 最终谁赢, 他最少出价多少.

set维护每个人最后一次投票的时间, 每次询问直接暴力找到最后一个未删除的, 假设为$x$, 那么$x$就是最后赢家, 求最少出价的话, 只要$x$的出价大于$x$之前一位的最大出价即可.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head


const int N = 1e6+10;
int m, n, a[N], b[N], pos[N], no[N];
set<int> c[N];

int main() {
	scanf("%d", &n);
	REP(i,1,n) {
		scanf("%d%d", a+i, b+i);
		pos[a[i]] = i;
		no[i] = a[i];
		c[a[i]].insert(b[i]);
	}
	set<int,greater<int> > q;
	REP(i,1,n) if (pos[i]) q.insert(pos[i]);
	scanf("%d", &m);
	REP(i,1,m) {
		int k, t;
		scanf("%d", &k);
		set<int> del;
		REP(i,1,k) scanf("%d", &t),del.insert(t);
		int x = 0;
		for (auto &&t:q) if (!del.count(no[t])) {
			x = no[t];
			del.insert(x);
			break;
		}
		if (!x) {puts("0 0"); continue;}
		int y = 0;
		for (auto &&t:q) if (!del.count(no[t])) {
			y = no[t];
			break;
		}
		if (!y) printf("%d %d
", x, *c[x].begin());
		else {
			int w = *(--c[y].end());
			printf("%d %d
", x, *c[x].lower_bound(w));
		}
	}
}