June 11th 模拟赛C T1 Sandcas Solution
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Description:
FJ居住的城堡有N个城墙(1<=N<=25,000),编号为1到N,每个城墙的高度为M_i(1<=M_i<=100,000),他想把城墙的高度按照某种顺序变成B_1,B_2,….B_N(1<=B_i<=100,000),由于工作量比较大,FJ雇佣了一些牛来帮忙提高或降低墙的高度,当然要付工资,提高一单位高度需要花费X(1<=X<=100),降低一单位高度需要花费Y(1<=Y<=100)。
FJ想用最少的钱来完成这项任务,请你帮助计算,保证答案在longint范围内。
Input
第1行:3个空格隔开的整数N,X和Y
第2到N+1行:第i+1行包含两个空格隔开的整数表示M_i和B_i
Output
输出一个整数表示最少花费。
Solution
将M与B排序,之后直接对应计算ANS。
Program
type QS=array [1..25000] of longint;var n,x,y,i,j:longint; ans:int64; m,b:QS;procedure swap(var x,y:longint);var t:longint;begin t:=x; x:=y; y:=t;end;procedure Qsort(var d:QS;i,j:longint);procedure sort(l,r:longint);var m,i,j:longint;begin i:=l; j:=r; m:=d[(i+j) shr 1]; repeat while d[i]<m do inc(i); while m<d[j] do dec(j); if i<=j then begin swap(d[i],d[j]); inc(i); dec(j); end; until i>j; if i<r then sort(i,r); if l<j then sort(l,j);end;begin sort(i,j);end;begin assign(input,'sandcas.in'); assign(output,'sandcas.out'); reset(input); rewrite(output); readln(n,x,y); for i:=1 to n do readln(m[i],b[i]); Qsort(m,1,n); Qsort(b,1,n); for i:=1 to n do if m[i]>b[i] then ans:=ans+(m[i]-b[i])*y else ans:=ans+(b[i]-m[i])*x; writeln(ans); close(input); close(output);end.
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