1/22 测试一（STL 模拟 贪心）C.（贪心，给出气球，输出最好成绩）Contest Balloons
1/22 测试一（STL 模拟 贪心）
C.（贪心，给出气球，输出最好成绩）Contest Balloons
One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn’t matter and teams are sorted only by the number of balloons they have. It means that one’s place is equal to the number of teams with more balloons, increased by 1. For example, if there are seven teams with more balloons, you get the eight place. Ties are allowed.
You should know that it’s important to eat before a contest. If the number of balloons of a team is greater than the weight of this team, the team starts to float in the air together with their workstation. They eventually touch the ceiling, what is strictly forbidden by the rules. The team is then disqualified and isn’t considered in the standings.
A contest has just finished. There are n teams, numbered 1 through n. The i-th team has ti balloons and weight wi. It’s guaranteed that ti doesn’t exceed wi so nobody floats initially.
Limak is a member of the first team. He doesn’t like cheating and he would never steal balloons from other teams. Instead, he can give his balloons away to other teams, possibly making them float. Limak can give away zero or more balloons of his team. Obviously, he can’t give away more balloons than his team initially has.
What is the best place Limak can get?
Input
The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team.
Output
Print one integer denoting the best place Limak can get.
Example
Input
8
20 1000
32 37
40 1000
45 50
16 16
16 16
14 1000
2 1000
Output
3
Input
7
4 4
4 4
4 4
4 4
4 4
4 4
5 5
Output
2
Input
7
14000000003 1000000000000000000
81000000000 88000000000
5000000000 7000000000
15000000000 39000000000
46000000000 51000000000
0 1000000000
0 0
Output
2
Note
In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1、Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak has only 14 balloons now and he would get the fifth place.
2、Limak gives 6 balloons away to a team with 45 balloons. Now they have 51 balloons and weight 50 so they fly and get disqualified.
3、Limak gives 1 balloon to each of two teams with 16 balloons initially.
4、Limak has 20 - 6 - 6 - 1 - 1 = 6 balloons.
5、There are three other teams left and their numbers of balloons are 40, 14 and 2.
6、Limak gets the third place because there are two teams with more balloons.
In the second sample, Limak has the second place and he can’t improve it.
In the third sample, Limak has just enough balloons to get rid of teams 2, 3 and 5 (the teams with 81 000 000 000, 5 000 000 000 and 46 000 000 000 balloons respectively). With zero balloons left, he will get the second place (ex-aequo with team 6 and team 7).
题意：
共有n个队伍，你是第一队的成员，每个队有t个气球，队伍的重量是w，若w > t ，则队伍会飞起来，失去比赛资格。你可以将气球给别的队伍，使他们飞起来，但是你不接受别的队伍的气球。要求输出你可以得到的最好名次。
思路：
将排在你前面的队伍单独拿出来，将他们距离飞起来需要的气球从小到大排序。每次取出飞起代价最小的那个，如果你能够让他飞起来，那么就让他飞起来，将他踢出队伍，同时要注意更新新的排序，将排在你前面的队伍放到前面的序列中，重复操作。在这个过程中不断更新取得的名次，最终取过程中获得的最小名次即可。
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <vector>
#include <map>
#include <queue>
#define MAXN 300005
#define INF 0x3f3f3f3f3f3f3f3f
typedef long long ll;
using namespace std;int main()
{int n;scanf("%d", &n);ll me = 0, t, w;int ans, cur = 1;priority_queue<ll, vector<ll>, greater<ll> > big;multimap<ll, ll, greater<ll> > sma;for(int i = 0; i < n; i++){scanf("%lld%lld", &t, &w);if(i == 0)me = t;else if(t > me){big.push(w - t + 1);
// ll p = big.top();
// cout<<'@'<<p<<endl;cur++;}else if(t <= me)sma.insert(pair<ll, ll>(t, w));}ans = cur;
// cout<<"ans: "<<ans<<endl;
// ll p = big.top();
// cout<<'*'<<p<<endl;while(!big.empty() && big.top() <= me){
// ll p = big.top();
// cout<<'*'<<p<<endl;me -= big.top();
// cout<<"me: "<<me<<endl;big.pop();cur--;multimap<ll, ll, greater<ll>> ::iterator iter = sma.begin();
// p = iter->first;
// cout<<'#'<<p<<endl;while(!sma.empty() && iter->first > me){cur++;
// p = iter->first;
// cout<<'#'<<p<<endl;big.push(iter->second - iter->first + 1);
// cout<<"charu"<<iter->second - iter->first + 1<<endl;sma.erase(iter);iter = sma.begin();}ans = min(ans, cur);
// cout<<"cur: "<<cur<<endl;
// cout<<"ans: "<<ans<<endl;}printf("%d\n", ans);return 0;
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