The War&&贪心算法

Description

A war had broken out because a sheep from your kingdom ate some grasses which belong to your neighboring kingdom. The counselor of your kingdom had to get prepared for this war. There areN (1 <= N <= 2500) unarmed soldier in your kingdom and there areM (1 <= M <= 40000) weapons in your arsenal. Each weapon has a weightW (1 <= W <= 1000), and for soldier i, he can only arm the weapon whose weight is betweenminWi and maxWi ( 1 <= minWi <= maxWi <= 1000). More armed soldier means higher success rate of this war, so the counselor wants to know the maximal armed soldier he can get, can you help him to win this war?

Input

There multiple test cases. The first line of each case are two integers N, M. Then the following N lines, each line contain two integers minWi, maxWi for each soldier. Next M lines, each line contain one integer W represents the weight of each weapon.

Output

For each case, output one integer represents the maximal number of armed soldier you can get.

Sample Input

3 31 53 75 104892 25 1010 20421

Sample Output

20

思路：按每个人所要求的武器的最大重量从小到大排序，再按给定的武器的重量从小到大排序，然后枚举所有人尽可能多的找满足要求的武器，

AC代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<string.h>#include<string>#include<set>using namespace std;typedef struct{int x;int y;}Node;Node s[2505];bool cmp(Node a,Node b){return a.y<b.y;}int main(){int  n,m;while(~scanf("%d%d",&n,&m)){for(int i=0;i<n;++i)scanf("%d%d",&s[i].x,&s[i].y);  sort(s,s+n,cmp);     multiset<int>Q;for(int i=1;i<=m;++i) {int a;scanf("%d",&a);Q.insert(a);}int ans=0;multiset<int>::iterator it;for(int i=0;i!=n;++i){it=Q.lower_bound(s[i].x);if(it!=Q.end()&&*it<=s[i].y) ans++,Q.erase(it);}printf("%d\n",ans);}return 0;}