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; } 
