今天是真的萎，切不动题，瞎写陈年老题找自信。。。
强行一波贪心猛如虎
先按毁坏时间排序，枚举，能修的就修，修不了就把前面耗时最长的拿出来和当前比较，假如现在需要时间更短就换
总之就是维护修了ans个的最快时间
#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>#include<set>using namespace std;struct node{ ???int t,ed;}a[210000];bool cmp(node n1,node n2){return n1.ed<n2.ed;}multiset<int> s;int main(){ ???int n; ???scanf("%d",&n); ???for(int i=1;i<=n;i++)scanf("%d%d",&a[i].t,&a[i].ed); ???sort(a+1,a+n+1,cmp); ???????int now=0,ans=0; ???for(int i=1;i<=n;i++) ???{ ???????if(a[i].t+now<=a[i].ed) ???????{ ???????????now+=a[i].t; ???????????ans++; ???????????s.insert(a[i].t); ???????} ???????else ???????{ ???????????int q=*--s.end(); ???????????if(a[i].t<q) ???????????{ ???????????????int sum=s.count(q); ???????????????s.erase(q); ???????????????for(int j=1;j<sum;j++)s.insert(q); ???????????????s.insert(a[i].t); ???????????????now+=a[i].t-q; ???????????} ???????} ???} ???printf("%d\n",ans); ???return 0;}bzoj1029: [JSOI2007]建筑抢修
原文地址：https://www.cnblogs.com/AKCqhzdy/p/8933430.html