《一生多事のFarmerJohn》

《一生多事のFarmerJohn》

这道题一看就是贪心题

先买便宜的牛奶，但如果牛奶的价钱一样，就先去买牛奶数量多的

这样写实在是太简洁了

尤其是这段代码

inline bool cmp(node x, node y){
     
	if(x.p != y.p) return x.p < y.p;
	return x.a > y.a;
}

#include
#define ll long long
#define maxn 5005
using namespace std;
inline int read(){
     
	int x = 0; bool f = false; char ch = getchar();
	while(!isdigit(ch)){
     if(ch == '-') f = true; ch = getchar();}
	while(isdigit(ch)){
     x = x * 10 + ch - '0'; ch = getchar();}
	return f ? -x : x;
}
inline void write(ll x){
     
	if (x < 0) x = ~x + 1, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
int n, m, ans;
struct node{
     
	int p, a;
}milk[maxn];
inline bool cmp(node x, node y){
     
	if(x.p != y.p) return x.p < y.p;
	return x.a > y.a;
}
int main(){
     
	n = read(), m = read();
	for(register int i = 1; i <= m; ++i) milk[i].p = read(), milk[i].a = read();
    sort(milk + 1, milk + m + 1, cmp);
    int i = 1;
    while(n){
     
    	if(milk[i].a){
     
    		--milk[i].a;
    		ans += milk[i].p;
    		--n;
		}
		else ++i;
	}
	write(ans);
	puts("");
	return 0;
}

也可以用队列＋$pair$

#include
#define ll long long
using namespace std;
inline int read(){
     
	int x = 0; bool f = false; char ch = getchar();
	while(!isdigit(ch)){
     if(ch == '-') f = true; ch = getchar();}
	while(isdigit(ch)){
     x = x * 10 + ch - '0'; ch = getchar();}
	return f ? -x : x;
}
inline void write(ll x){
     
	if (x < 0) x = ~x + 1, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
priority_queue<pair<int, int> > q;
int n, m, ans, p, a, milk_quantity;
int main(){
     
	n = read(), m = read();
	if(!n){
     puts("0"); return 0;}
	for(register int i = 1; i <= m; ++i){
     
		p = read(); a = read();
		q.push(make_pair(-p, a));//小技巧:存负边,保证取出的是最小值,你愿意也可以写个运算重载符
	}
	while(true){
     
		if(milk_quantity + q.top().second >= n){
     
			ans += (n - milk_quantity) * (-q.top().first);//注意存的是负边,取时要变正号,下同
			write(ans);
			puts("");
			return 0;
		}
		milk_quantity += q.top().second;
		ans += q.top().second * (-q.top().first);
		q.pop(); 
	}
}