UVa 1346 Songs (greedy good question)

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem &problem=4092 thought: Directly look at this and the formula, remember a (i) =f (i) Σl (j) (1<=j<=i), then Sum=σa (i) (not to be confused by subscript, S (i) can be completely replaced by i).

Set sum has reached the minimum,

Note Sum1 = A (i) +a (i+1) = (f (i) +f (i+1)) * (L (1) +...+l (i-1)) + f (i) *l (i) + f (i+1) *l (i) + f (i+1) *l (i+1)

Swap f (i) and F (i+1), Exchange L (i) and L (i+1), sum2= (f (i) +f (i+1)) * (L (1) +...+l (i-1)) + f (i) *l (i) + f (i) *l (i+1) + f (i+1) *l (i+1)

Because the sum is the smallest, there must be sum1<=sum2, so f (i+1) *l (i) <=f (i) *l (i+1)

(It can also be understood that the greater the frequency, the shorter the length, that is, f (i)/L (i) the larger the position, the more forward)

Complete code:

01./*0.032s*/
.  
#include <bits/stdc++.h>  
04.using namespace std;  
06.struct node  
07.{  
an    int id;    double L, F;    BOOL Operator < (const node& a) const
.    {a  
. return        A.F * L < F * A.L;  
a[65540];  
16.int Main ()  
17.{    int n, I, S;  
While    (~scanf ("%d", &n))  
.    {for        (i = 1; I <= n; ++i)  
.            scanf ("%d%lf%lf", &a[i].id, &A[I].L, &A[I].F);        scanf ("%d", &s);        Nth_element (A + 1, a + s, a + n + 1);        printf ("%d\n", a[s].id);  
.    return 0;  
28.}

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