測试赛D - The War（有控制范围的贪心）

D - The War

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

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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 are N (1 <= N <= 2500) unarmed soldier in your kingdom and there are M (1 <= M <= 40000) weapons in your arsenal. Each weapon has a weight W (1 <= W <= 1000), and for soldier i, he can only arm the weapon whose weight is between minWi 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 3
1 5
3 7
5 10
4
8
9
2 2
5 10
10 20
4
21

Sample Output

2
0

贪心的问题，人能够使用的有一些范围，武器有重量，唯一要解决的问题就是在一个人的控制范围内包括还有一个人的，这样的情况要让范围小的先找

y有小到大，假设y同样时x由小到大。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct node{
    int x , y ;
} p[2600];
int q[1200] ;
bool cmp(node a,node b)
{
    return a.y < b.y || ( a.y == b.y && a.x < b.x ) ;
}
int main()
{
    int i , j , n , m ;
    while(scanf("%d %d", &n, &m) !=EOF )
    {
        memset(q,0,sizeof(q));
        for(i = 0 ; i < n ; i++)
            scanf("%d %d", &p[i].x, &p[i].y);
        while(m--)
        {
            scanf("%d", &i);
            q[i]++ ;
        }
        m = 0 ;
        sort(p,p+n,cmp);
        for(i = 0 ; i < n ; i++)
        {
            int l = p[i].x , r = p[i].y ;
            for(j = l ; j <= r ; j++)
                if( q[j] )
            {
                q[j]--;
                m++ ;
                break;
            }
        }
        printf("%d\n", m);
    }
    return 0 ;
}