The War

The War

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 31 53 75 104892 25 1010 20421

Sample Output

20

题意为给定一个士兵所承受的重量范围，让其挑选兵器。

将承重范围和兵器的重量都进行从小到大的排序，再用两个循环进行筛选

<span style="font-size:18px;color:#33cc00;">#include <iostream>#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>using namespace std;struct node{    int u,v;}zhong[2511];int cmp(struct node a,struct node b){    return a.v<b.v;}int main(){    int n,m,i,j,num;    int a[40014];    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=0;i<n;i++)        scanf("%d%d",&zhong[i].u,&zhong[i].v);        for(i=0;i<m;i++)        scanf("%d",&a[i]);        sort(zhong,zhong+n,cmp);        sort(a,a+m);        num=0;        for(i=0;i<m;i++)        {            for(j=0;j<n;j++)            {                if(a[i]>=zhong[j].u&&a[i]<=zhong[j].v&&a[i]!=-1&&zhong[j].v!=-1)                {                    num++;                    a[i]=-1;                    zhong[j].v=-1;                }            }        }        printf("%d\n",num);    }}</span>