题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4864
Total Submission(s): 1035 Accepted Submission(s): 245
The company has n machines. Each machine has a maximum working time and a level. If the time for the task is more than the maximum working time of the machine, the machine can not complete this task. Each machine can only complete a task one day. Each task can only be completed by one machine.
The company hopes to maximize the number of the tasks which they can complete today. If there are multiple solutions, they hopes to make the money maximum.
The first line contains two integers N and M. N is the number of the machines.M is the number of tasks(1 < =N <= 100000,1<=M<=100000).
The following N lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the maximum time the machine can work.yi is the level of the machine.
The following M lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the time we need to complete the task.yi is the level of the task.
那么我们首先先对任务排序,先按时间排序,若时间一样,则依照level排。
之后呢,我们依次选择时间和level最接近该任务的就可以。
因为这道题数据范围比較大,在tle了n次之后,所以我參考了某神牛的博客,採用了set和二分查找进行优化,时间复杂度瞬间就降下来了。。。
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<cstdlib> #include<vector> #include<set> #include<map> #include<bitset> using namespace std; typedef long long ll; const int MAX=100005; struct Task{ int x,y; }t[MAX]; bool cmp(Task a,Task b){if(a.x==b.x) return a.y>b.y;else return a.x>b.x;}; inline ll f(int x,int y){return 500*x+2*y;} int main(){ int n,m; while(cin>>n>>m){ multiset<int> st[105]; int maxlevel=0,cnt=0; ll sum=0; for(int i=0;i<n;i++){ int x,y; scanf("%d%d",&x,&y); st[y].insert(x); maxlevel=max(maxlevel,y); } for(int i=0;i<m;i++){scanf("%d%d",&t[i].x,&t[i].y);} sort(t,t+m,cmp); for(int i=0;i<m;i++){ int u=t[i].x,v=t[i].y; for(int j=v;j<=maxlevel;j++){ if(st[j].empty()) continue; multiset<int>::iterator it=st[j].lower_bound(u); if(it==st[j].end()||*it<u) continue; else{ cnt++;sum+=f(u,v); st[j].erase(it); break; } } } cout<<cnt<<" "<<sum<<endl; } return 0; }