题目链接:https://www.acwing.com/problem/content/147/
贪心策略:
将商品按时间排序,维护一个小根堆,如果当前商品过期天数等于堆中商品数量,
且当前商品价值大于堆顶商品,则将堆顶弹出,插入当前商品;
如果当前商品过期天数大于堆中商品数量,则之间将商品插入堆中
最后统计堆中所有商品价值之和
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 100010;
int n, tot, ans;
struct Object{
int p, d;
}a[maxn];
bool cmp(Object x, Object y){ return x.d < y.d; }
int heap[maxn];
void up(int pos){
while(pos > 1){
if(heap[pos] < heap[pos / 2]) {
swap(heap[pos], heap[pos / 2]);
pos /= 2;
}else break;
}
}
void insert(int x){
heap[++tot] = x;
up(tot);
}
void down(int pos){
int s = pos * 2;
while(s <= tot){
if(s < tot && heap[s] > heap[s + 1]) ++s;
if(heap[s] < heap[pos]){
swap(heap[s] , heap[pos]);
pos = s, s = pos * 2;
} else break;
}
}
void extract(){
heap[1] = heap[tot--];
down(1);
}
ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }
int main(){
while(scanf("%d",&n) != EOF){
tot = 0; ans = 0;
for(int i=1;i<=n;++i){
scanf("%d%d",&a[i].p, &a[i].d);
}
sort(a + 1, a + 1 + n, cmp);
for(int i=1;i<=n;++i){
if(a[i].d > tot){
insert(a[i].p);
} else if(a[i].d == tot){
if(a[i].p > heap[1]){
extract();
insert(a[i].p);
}
}
}
for(int i=1;i<=tot;++i){
ans += heap[i];
}
printf("%d
",ans);
}
return 0;
}