zoukankan      html  css  js  c++  java
  • poj_1201_Intervals

    Description

    You are given n closed, integer intervals [ai, bi] and n integers c1, …, cn.
    Write a program that:
    reads the number of intervals, their end points and integers c1, …, cn from the standard input,
    computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,…,n,
    writes the answer to the standard output.

    Input

    The first line of the input contains an integer n (1 <= n <= 50000) – the number of intervals. The following n lines describe the intervals. The (i+1)-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50000 and 1 <= ci <= bi - ai+1.

    Output

    The output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each i=1,2,…,n.

    题解

    可以化为一类线性不等式定义的线性函数。而这类问题其实是可以转化为单元最短路径问题,从而用刚才所准备到的Bellman Ford算法来解决它。

    代码

    type
      arr=record
            x,y,w:longint;
          end;
    var
      d:array [0..150001] of longint;
      a:array [0..150001] of arr;
      n,nm,min,max:longint;
    procedure add(u,v,z:longint);
    begin
      inc(nm);
      with a[nm] do
        begin
          x:=u; y:=v; w:=z;
        end;
    end;
    
    procedure init;
    var
      i,x,y,z:longint;
    begin
      readln(n); nm:=0;
      max:=0; min:=maxlongint;
      for i:=1 to n do
        begin
          readln(x,y,z); dec(x);
          add(y,x,-z);
          if max<y then max:=y;
          if min>x then min:=x;
        end;
      for i:=min+1 to max do
        begin
          add(i-1,i,1);
          add(i,i-1,0);
        end;
    end;
    
    procedure main;
    var
      i:longint;
      bo:boolean;
    begin
      fillchar(d,sizeof(d),63);
      while 1=1 do
        begin
          bo:=true;
          for i:=1 to nm do
            with a[i] do
              if d[y]>d[x]+w then
                begin
                  d[y]:=d[x]+w; bo:=false;
                end;
          if bo then break;
        end;
    end;
    
    begin
      init;
      main;
      write(d[max]-d[min]);
    end.
    
  • 相关阅读:
    飞鸽传书中文源码
    nohup命令参考
    Linux平台编程新手入门 C语言中的移位操作
    小技巧:让linux程序在后台运行
    2440之中断管理
    linux终端中输出彩色字体(C/SHELL)
    C语言标准中的逻辑位移和算术位移
    SQL2005利用ROW_NUMER实现分页的两种常用方式
    不用现有方法,把string转换成int型[C#]
    C# 如何生成一个时间戳
  • 原文地址:https://www.cnblogs.com/zyx-crying/p/9319541.html
Copyright © 2011-2022 走看看