CodeForces 1379C : Choosing flowers 二分

传送门

题目描述

你想买n朵花,商店提供了m种,第i种花,买第一支可以获得a[i]的价值,之后无论卖多少支,都是b[i]的价值,每种花都无限提供,问你能获得的最大价值是多少?

分析

挺水的一道题,就是细节挺多的
首先,最多只有一朵花会被购买一次以上,这个应该比较好证明,然后我们去枚举被多次购买的花,如果a[j] > b[i],说明j朵花第一次也需要被购买,排个序二分查找一下范围即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
const ll mod= 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a){
    
    char c=getchar();T x=0,f=1;while(!isdigit(c)){
    
    if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){
    
    x=(x<<1)+(x<<3)+c-'0';c=getchar();}a=f*x;}
int gcd(int a,int b){
    
    return (b>0)?gcd(b,a%b):a;}
ll n,m;
struct Node{
    
    
    ll x,y;
}a[N];
ll w[N],sum[N];

bool cmp(Node A,Node B){
    
    
    if(A.x != B.x) return A.x < B.x;
    else return A.y > B.y;
}

int main(){
    
    
    int T;
    read(T);
    while(T--){
    
    
        memset(w,0,sizeof w);
        memset(sum,0,sizeof sum);
        read(m),read(n);
        for(ll i = 1;i <= n;i++){
    
    
            read(a[i].x),read(a[i].y);
        }
        sort(a + 1,a + 1 + n,cmp);
        for(ll i = 1;i <= n;i++) {
    
    
            w[i] = a[i].x;
            sum[i] = sum[i - 1] + w[i];
        }
        ll ans = 0;
        for(ll i = 1;i <= n;i++){
    
    
            ll t = a[i].y;
            ll p = upper_bound(w + 1,w + 1 + n,t) - w;
            ll res;
            if(w[p] > t){
    
    
                if(n - p + 1 >= m){
    
    
                    res = sum[n] - sum[n - m];
                }
                else{
    
    
                    res = sum[n] - sum[p - 1];
                    res = res + t * (m - (n - p + 1) - 1);
                    if(p <= i) res += t;
                    else res += a[i].x;
                }
            }
            else{
    
    
                res = a[i].x + (m - 1) * t;
            }
            ans = max(ans,res);
        }
        if(m == 1) ans = sum[n] - sum[n - 1];
        dl(ans);
    }
    return 0;
}

/**
*  ┏┓   ┏┓+ +
* ┏┛┻━━━┛┻┓ + +
* ┃       ┃
* ┃   ━   ┃ ++ + + +
*  ████━████+
*  ◥██◤ ◥██◤ +
* ┃   ┻   ┃
* ┃       ┃ + +
* ┗━┓   ┏━┛
*   ┃   ┃ + + + +Code is far away from  
*   ┃   ┃ + bug with the animal protecting
*   ┃    ┗━━━┓ 神兽保佑,代码无bug 
*   ┃        ┣┓
*    ┃        ┏┛
*     ┗┓┓┏━┳┓┏┛ + + + +
*    ┃┫┫ ┃┫┫
*    ┗┻┛ ┗┻┛+ + + +
*/





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