Educational Codeforces Round 43 (Rated for Div. 2) E. Well played!

最新推荐文章于 2020-07-26 10:13:35 发布

原创最新推荐文章于 2020-07-26 10:13:35 发布 · 241 阅读

0 ·

CC 4.0 BY-SA版权

竞赛专栏收录该内容

87 篇文章

订阅专栏

本文介绍了一种在游戏“BrainStone”中优化生物单位伤害输出的方法。玩家通过使用有限次数的两种魔法技能来最大化所有生物单位的总伤害值。文章详细解释了问题背景，并提供了一段C++代码实现解决方案。

Recently Max has got himself into popular CCG "BrainStone". As "BrainStone" is a pretty intellectual game, Max has to solve numerous hard problems during the gameplay. Here is one of them:

Max owns n creatures, i-th of them can be described with two numbers — its health hpi and its damage dmgi. Max also has two types of spells in stock:

Doubles health of the creature (hpi := hpi·2);
Assigns value of health of the creature to its damage (dmgi := hpi).

Spell of first type can be used no more than a times in total, of the second type — no more than b times in total. Spell can be used on a certain creature multiple times. Spells can be used in arbitrary order. It isn't necessary to use all the spells.

Max is really busy preparing for his final exams, so he asks you to determine what is the maximal total damage of all creatures he can achieve if he uses spells in most optimal way.

Input

The first line contains three integers n, a, b (1 ≤ n ≤ 2·105, 0 ≤ a ≤ 20, 0 ≤ b ≤ 2·105) — the number of creatures, spells of the first type and spells of the second type, respectively.

The i-th of the next n lines contain two number hpi and dmgi (1 ≤ hpi, dmgi ≤ 109) — description of the i-th creature.

Output

Print single integer — maximum total damage creatures can deal.

Examples
inputCopy
2 1 1
10 15
6 1
outputCopy
27
inputCopy
3 0 3
10 8
7 11
5 2
outputCopy
26

Note

In the first example Max should use the spell of the first type on the second creature, then the spell of the second type on the same creature. Then total damage will be equal to 15 + 6·2 = 27.

In the second example Max should use the spell of the second type on the first creature, then the spell of the second type on the third creature. Total damage will be equal to 10 + 11 + 5 = 26.

题解：
最优解一定是对同一个数连续乘2
code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=200005;
struct node
{
    ll x,y;
}c[maxn];
bool cmp(const node &a,const node &b)
{
    return a.x-a.y>b.x-b.y;
}
int main()
{
    ll n,a,b;
    scanf("%lld%lld%lld",&n,&a,&b);
    ll ans=1;
    while(a--)ans*=2;
    for(int i=1;i<=n;i++)
        scanf("%lld%lld",&c[i].x,&c[i].y);
    sort(c+1,c+n+1,cmp);
    ll sum=0;
    for(int i=1;i<=min(n,b);i++)
    {
        sum+=max(c[i].x,c[i].y);
    }
    for(int i=min(n,b)+1;i<=n;i++)
        sum+=c[i].y;
    ll ma=sum;
    if(ans==1||b==0)
    {
        printf("%lld\n",ma);
        return 0;
    }
    for(int i=1;i<=min(n,b);i++)
    {
        ll mm=sum-max(c[i].x,c[i].y)+c[i].x*ans;
        ma=max(ma,mm);
    }
    sum-=max(c[min(n,b)].x,c[min(n,b)].y);
    sum+=c[min(n,b)].y;
    for(int i=min(n,b)+1;i<=n;i++)
    {
        ll mm=sum-c[i].y+c[i].x*ans;
        ma=max(ma,mm);
    }
    printf("%lld\n",ma);
    return 0;
}