Codeforces 725D[Contest Balloons]【贪心】

Description

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?

题解

如果我要把气球分给别人，那么我一定是优先分给那些排在我前面的，并且需要最少的气球就可以上天的，并且，我每次分掉一些气球后会有另一些人排到我的前面去，所以，只要用一个小根堆维护一下，从每次的答案中刷出最小值就是答案。

代码

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#define maxn 300006#define LL long longusing namespace std;inline char nc(){    static char buf[100000],*i=buf,*j=buf;    return i==j&&(j=(i=buf)+fread(buf,1,100000,stdin),i==j)?EOF:*i++;}inline LL _read(){    char ch=nc();LL sum=0;    while(!(ch>='0'&&ch<='9'))ch=nc();    while(ch>='0'&&ch<='9')sum=sum*10+ch-48,ch=nc();    return sum;}struct data{    LL t,w;int id;    bool operator >(const data&b)const{return w-t>b.w-b.t;};}a[maxn];int n,ans;priority_queue<data,vector<data>,greater<data> >heap;bool cmp(data x,data y){    return x.t>y.t||(x.t==y.t&&x.id<y.id);}int main(){    freopen("balloons.in","r",stdin);    freopen("balloons.out","w",stdout);    n=_read();    for(int i=1;i<=n;i++)a[i].t=_read(),a[i].w=_read()+1,a[i].id=i;    sort(a+1,a+1+n,cmp);    int p;    for(int i=1;i<=n;i++) if(a[i].id==1){        p=i;break;    }    for(int i=1;i<p;i++)heap.push(a[i]);    ans=p-1;    int num=p-1,j=p+1;    while(!heap.empty()&&a[p].t){        data c=heap.top();        if(a[p].t>=c.w-c.t){            heap.pop();num--;            a[p].t-=c.w-c.t;            while(j<=n&&a[j].t>a[p].t){                heap.push(a[j++]);                num++;            }            ans=min(ans,num);        }else break;    }    printf("%d",ans+1);    return 0;}