POJ 3614

It is greedy and priority queues, very smart to solve: 

　　This greed is very clever ah

 

 
C, L;
 const  int MAX_C = 2500 ;
 const  int MAX_L = 2500 ; 
P A [MAX_L];          // store sunscreen 
P B [MAX_C];          // for storing bovine
 // qsort ordering function 
int CMP ( const  void A *, const  void * B) { 
    P * AA = (P * ) A; 
    P * BB = (P * ) B;
     return AA-> First> BB-> First? . 1 : - . 1 ;       // the SPF is ordered Preferences 
} 

int  main ()
{
    cin>>C>>L;
    P temp;
    priority_queue<int, vector<int>, greater<int> > pque;             //注意是由大到小出现的
    for(int i = 0; i < C; i++){
        scanf("%d%d", &b[i].first, &b[i].second);
    }
    for(int i = 0; i < L; i++){
        scanf("%d%d", &a[i].first, &a[i].second);
    }
    //升序排序， 完成对优先队列的存储
    qsort(a, L, sizeof(a[0]), cmp);
    qsort(b, C, sizeof(b[0]), cmp);
   /* printf("SHOW THE BOTTLE!\n");
    for(int i = 0; i < L; i++)  printf("%d %d\n", a[i].first, a[i].second);   puts("");
    printf("SHOW THE COW : \n");
    for(int i = 0; i < C; i++){
        printf("%d  %d\n", b[i].first, b[i].second);
    }*/
    int res = 0;    int j = 0;
    0I =int(forwe must pay attention to the use of a cream i; b cattle, using J//
    ; I <L; I ++) {      // array sunscreen traverse 
        the while (J <&& A C [I] .first> = B [J] .first) { 
            pque.push (B [J] .second) ; // maximum value of cattle, priority queue, because he's the minimum (section head) must be satisfied, we see here is the key to his maximum (tail section) 
            J ++ ; 
        } 
        the while (pque!. empty () && A [I] .second) {     // queue is not empty, sunscreen unspent 
            int X = pque.top (); pque.pop (); // priority queue been taken out 
            IF (X <A [ I] .first)   Continue ;    // head meet, but do not satisfy the tail section 
            RES ++; a [I] .second--;   // for answer, and update the number of sunscreen 
        } 
    } 
    the printf ( "%d\n", res);
    return 0;
}