Leaving Auction CodeForces - 749D (set,贪心,模拟)

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 '\n'

#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\n", x, *c[x].begin());

		else {

			int w = *(--c[y].end());

			printf("%d %d\n", x, *c[x].lower_bound(w));

		}

	}

}