1634: [Usaco2007 Jan]Protecting the Flowers 护花

Farmer John went to cut some wood and left N (2 <= N <= 100,000) cows eating the grass, as usual. When he returned, he found to his horror that the cows were in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport the cows back to their barn. Each cow i is at a location that is Ti minutes (1 <= Ti <= 2,000,000) away from the barn. Furthermore, while waiting for transport, she destroys Di (1 <= Di <= 100) flowers per minute. No matter how hard he tries,FJ can only transport one cow at a time back to the barn. Moving cow i to the barn requires 2*Ti minutes (Ti to get there and Ti to return). Write a program to determine the order in which FJ should pick up the cows so that the total number of flowers destroyed is minimized.

* Line 1: A single integer

N * Lines 2..N+1: Each line contains two space-separated integers, Ti and Di, that describe a single cow‘s characteristics

    第1行输入N，之后N行每行输入两个整数Ti和Di．

* Line 1: A single integer that is the minimum number of destroyed flowers

    一个整数，表示最小数量的花朵被吞食．

    约翰用6，2，3，4，1，5的顺序来运送他的奶牛．

Silver

 题解：呵呵呵，又忘开int64了，结果WA了一次（长点记性吧QAQ）。。。我们可以这样分析此题——如题，假如按照T_i,D_i,T_i+1,D_i+1的顺序为最优解的话，则此时消耗的花为2T_iD_i+1,假如将这两个的顺序反过来的话，则消耗的花为2T_i+1D_i，由于前者为最优解，所以T_iD_i+1<T_i+1D_i，即(T_i/D_i)<(T_i+1/D_i+1)，则可以用冒泡排序说明顺序中各个牛的(T_i/D_i)必然递增（虽是这么分析的，但还是不建议先做商再排序，小心卡精度）。。。That‘s all...

 1 var
 2    i,k,m,n:longint;
 3    j,l:int64;
 4    a,b:array[0..200000] of int64;
 5 procedure swap(var x,y:int64);
 6           var z:int64;
 7           begin
 8                z:=x;x:=y;y:=z;
 9           end;
10 procedure sort(l,r:longint);
11           var
12              i,j:longint;
13              x,y:int64;
14           begin
15                i:=l;j:=r;
16                x:=a[(l+r) div 2];
17                y:=b[(l+r) div 2];
18                repeat
19                      while (a[i]*y)<(x*b[i]) do inc(i);
20                      while (a[j]*y)>(x*b[j]) do dec(j);
21                      if i<=j then
22                         begin
23                              swap(a[i],a[j]);
24                              swap(b[i],b[j]);
25                              inc(i);dec(j);
26                         end;
27                until i>j;
28                if l<j then sort(l,j);
29                if i<r then sort(i,r);
30           end;
31 begin
32      readln(n);
33      for i:=1 to n do
34          begin
35               readln(a[i],b[i]);
36               a[i]:=a[i]*2;
37          end;
38      sort(1,n);
39      j:=0;l:=0;
40      for i:=1 to n-1 do
41          begin
42               j:=j+a[i];
43               l:=l+j*b[i+1];
44          end;
45      writeln(l);
46 end.