【HDU 4864】 Task

【题目链接】

http://acm.hdu.edu.cn/showproblem.php?pid=4864

【算法】

贪心

不妨将两个数组分别按x从大到小排序

然后枚举每件物品，选择x值大于该物品的且y值最小的机器对它进行修理

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100010
#define MAXP 110

int n,m,i,j,pos;
long long ans,money;
int cnt[MAXP];

struct info
{
        long long x,y;
} a[MAXN],b[MAXN];

inline bool cmp(info a,info b)
{
        return (a.x != b.x) ? a.x > b.x : a.y > b.y;
}

int main()
{

        while (scanf("%d%d",&n,&m) != EOF)
        {
                money = ans = ;
                memset(cnt,,sizeof(cnt));
                for (i = ; i <= n; i++) scanf("%lld%lld",&a[i].x,&a[i].y);
                for (i = ; i <= m; i++) scanf("%lld%lld",&b[i].x,&b[i].y);
                sort(a+,a+n+,cmp);
                sort(b+,b+m+,cmp);
                pos = ;
                for (i = ; i <= m; i++)
                {
                        while (pos <= n && a[pos].x >= b[i].x)
                        {
                                cnt[a[pos].y]++;
                                pos++;
                        }
                        for (j = b[i].y; j <= MAXP; j++)
                        {
                                if (cnt[j])
                                {
                                        cnt[j]--;
                                        ans++;
                                        money +=  * b[i].x +  * b[i].y;
                                        break;
                                }
                        }
                }
                printf("%lld %lld\n",ans,money);
        }

        return ;

}