推销员

题目描述

输入

输出

样例输入

Sample Input1:

5
1 2 3 4 5
1 2 3 4 5

Sample Input2:

5
1 2 2 4 5
5 4 3 4 1

样例输出

Sample Output1:

15
19
22
24
25

Sample Output2:

12
17
21
24
27

数据范围限制

提示

思路：

先打个暴力（60分）

然后就会发现，如果当前已经走到最后一个了，那么接下来的推销顺序就已经确定了

我们可以快排一下，然后输出

这虽然是水法，但却很有意义……
（正解是线段树……某位大犇考场AC……）

代码：

var        n,i,j,k,x,y,z,s,ls:longint;        a:array[0..100001,1..2]of longint;        b:array[0..100001]of boolean;procedure qs(l,r:longint);var        i,j,x:longint;        y:array[1..2]of longint;begin        i:=l;        j:=r;        x:=a[(l+r) div 2,1];        repeat                while a[i,1]<x do inc(i);                while x<a[j,1] do dec(j);                if not(i>j) then                begin                        y:=a[i];                        a[i]:=a[j];                        a[j]:=y;                        inc(i);                        dec(j);                end;        until i>j;        if l<j then qs(l,j);        if i<r then qs(i,r);end;procedure qs1(l,r:longint);var        i,j,x:longint;        y:array[1..2]of longint;        c:boolean;begin        i:=l;        j:=r;        x:=a[(l+r) div 2,2];        repeat                while a[i,2]>x do inc(i);                while x>a[j,2] do dec(j);                if not(i>j) then                begin                        y:=a[i];                        a[i]:=a[j];                        a[j]:=y;                        c:=b[i];                        b[i]:=b[j];                        b[j]:=c;                        inc(i);                        dec(j);                end;        until i>j;        if l<j then qs1(l,j);        if i<r then qs1(i,r);end;procedure qs2(l,r:longint);var        i,j,x:longint;        y:array[1..2]of longint;        c:boolean;begin        i:=l;        j:=r;        x:=a[(l+r) div 2,2];        repeat                while a[i,2]<x do inc(i);                while x<a[j,2] do dec(j);                if not(i>j) then                begin                        y:=a[i];                        a[i]:=a[j];                        a[j]:=y;                        c:=b[i];                        b[i]:=b[j];                        b[j]:=c;                        inc(i);                        dec(j);                end;        until i>j;        if l<j then qs2(l,j);        if i<r then qs2(i,r);end;begin        assign(input,'salesman.in');        reset(input);        assign(output,'salesman.out');        rewrite(output);        readln(n);        for i:=1 to n do        read(a[i,1]);        for i:=1 to n do        read(a[i,2]);        qs(1,n);        for i:=1 to n do        if ls<>a[i,1] then        begin                if i<>1 then qs2(x,i-1);                x:=i;                ls:=a[i,1];        end;        qs2(x,n);        ls:=0;        x:=0;        for i:=1 to n do        begin                for j:=1 to n do                if b[j]=false then                begin                        z:=0;                        if ls<j then z:=(a[j,1]-a[ls,1])*2;                        inc(z,a[j,2]);                        if z>=x then                        begin                                y:=j;                                x:=z;                        end;                end;                b[y]:=true;                if ls<y then ls:=y;                inc(s,x);                writeln(s);                if ls=n then                begin                        qs1(1,n);                        for j:=1 to n do                        if b[j]=false then                        begin                                inc(s,a[j,2]);                                writeln(s);                        end;                        halt;                end;                x:=0;        end;end.