学步园

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?

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.

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

3 3
1 5
3 7
5 10
4
8
9
2 2
5 10
10 20
4
21

2
0

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int maxx[5000];
int minn[5000];
int w[50000];
int main()
{
    int n,m;
    int i,j;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0; i<n; i++)
        {
            scanf("%d%d",&minn[i],&maxx[i]);
        }

        for(i=0; i<m; i++)
        {
            scanf("%d",&w[i]);
        }
        int t;
        for(i=0; i<n; i++)
        {
            for(j=0; j<n-1-i; j++)
            {
                if(maxx[j]>=maxx[j+1])
                {
                    t  = maxx[j];
                    maxx[j] = maxx[j+1];
                    maxx[j+1] = t;
                    t = minn[j];
                    minn[j] = minn[j+1];
                    minn[j+1] = t;
                }
            }
        }
        sort(w,w+m);
        int ans=0;
        for(i=0; i<m; i++)
        {
            for(j=0; j<n; j++)
            {
                if(w[i]<=maxx[j]&&w[i]>=minn[j])
                {
                    ans++;
                    maxx[j]=0;
                    minn[j]=0;
                    w[i]=0;
                    break;
                }

            }
        }
        printf("%d\n",ans);
    }
    return 0;
}