題目鏈接 4486 Task

按照時間從大到小排序 然後枚舉所有的y值 用一個數組存儲 符合要求就算上

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define maxn 200001
struct ac{
  LL  x,y;
}a[maxn],b[maxn];
LL  c[maxn];
bool cmp(ac q,ac w){
   if(q.x==w.x) return q.y>w.y;
   return q.x>w.x;
}
int main(){
   LL  n,m;
   
 
while(cin>>n>>m&&n&&m){
     for(LL  j=1;j<=n;j++){
        scanf("%lld%lld",&a[j].x,&a[j].y);
     }
     for(LL  j=1;j<=m;j++){
        scanf("%lld%lld",&b[j].x,&b[j].y);
     }
     sort(a+1,a+1+n,cmp);
     sort(b+1,b+1+m,cmp);
     LL  l
 
=1;
     LL ans=0,sum=0;
     memset(c,0,sizeof(c));
     for(LL  j=1;j<=m;j++){
        while(l<=n&&a[l].x>=b[j].x){
            c[a[l].y]++;
            l++;
        }
        for(LL k=b[j].y;k<=100;k++){
           if(c[k]){
              c[k]--;
              sum++;
              ans
 
+=1LL*(1LL*500*b[j].x+1LL*2*b[j].y);
              break;
           }
        }
     }
     cout<<sum<<" "<<ans<<endl;
   }
   return 0;
}

HDU--4486 Task（貪心）