codeforces-387B. George and Round

codeforces-387B. George and Round

题意：

某人G准备了m个题目，难度分别为b1,b2,b3,,,,,bm

他需要使用n个题目，要求难度分别为 a1,a2,a3,an

他有“特异功能”，可以使任意一个题目减低任意难度

给出n，m

【a】【b】

问：使用他准备的题目，同时使用“特异功能”，他还要再准备多少题目

题意抽象：

1、我们尽量使用他所准备的题目来形成n个题目

2、对于所有能直接使用的b[i]==a[j](1<=i<=m,1<=j<=n)

3、对于剩下的题目，降低难度使得b[x]-c==a[y](1<=x<=m,1<=y<=n)

4、对所有没有对应的准备好的a[u]计数，答案就是其总数

解题思路：

1、用map<int ,int >存储<a[i],x>――x是难度为C的题目的个数

2、输入b[x]的同时，在【a】中检索b[i]==a[j](1<=i<=m,1<=j<=n)，有的话，使用，否则存储到优先队列pb中

3、检索map<int ,int >，将未准备的题目的难度值取出，存储到优先队列pa中

4、将所准备的题目进行适当的降低难度处理，能用的就用

5，收集答案，输出

提交结果：

6088343

Feb 23, 2014 9:48:30 AM

20114045007

387B - George and Round

GNU C++

Accepted

15 ms

400 KB

代码：

#include<iostream> #include<map> #include<queue> using namespace std; int main(){     int n,m; //The number of the problem     int x;     map<int,int> ma;  //to select the prepare problem     priority_queue<int> pa,pb;  //to select the simplyfiy problem     cin>>n>>m;     while(n--)cin>>x,ma[x]++;  //save the problem information he need     while(m--){         cin>>x;         if(ma[x]>0)ma[x]--;  //if the problem he prepare that he need we use it         else pb.push(x);   //else we save the problem he prepare but we not used     }     map<int,int>::iterator it=ma.begin();  //to  serach the problem information he need but he has not prepared     while(it!=ma.end()){         while(it->second--)pa.push(it->first);   //to save it         it++;     }     while(!pa.empty()&&!pb.empty()){         if(pa.top()<pb.top())pa.pop(),pb.pop();  //if a problem he has prepare can be simpylify to use we use it         else pa.pop(),n++; //the problem he need ,but he not prepare and can't simpylify other problem for it we let him prepare agin     }     cout<<n+pa.size()+1; //why "-1",before cout the problem he hava to prepare agin n=-1     //pa.size(),,,the problem he need ,but he not prepare and can't simpylify other problem for it we let him prepare agin     return 0; }