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.
Sample Input

5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1
Sample Output

6
Source

Southwestern Europe 2002

分析:
差分约束系统模版题。
d[i]表示左边界到i的个数。
设读入x,y,w。
关系为y-x>=w,也就是说x-y<=-w,x到y有一条-w的边。
然后我们知道,
0<=d[i]-d[i-1]<=1（自己脑补）
也就是
d[i]-d[i-1]<=1,d[i-1]-d[i]<=0
就是说i和i-1有一条1的边，i-1和i有一条0的边，跑最短路。（这题本来为最长路，强行写成最短路的话要输出相反数）。

代码：

const  MaxE=500001;  MaxV=150001;type  rec=record   x,y,w,next:longint;  end;var  n,m,c,qx,i,x,y,w,o,q,maxx,maxy:longint;  g:array [0..Maxv] of rec;  ls:array [0..Maxe] of longint;  v,d,list,sum:array [0..maxe] of longint;procedure spfa(first:longint);var  head,tail,t,i,qe:longint;begin  tail:=1;  list[1]:=first;  for i:=maxx-1 to maxy do  d[i]:=maxlongint;  d[first]:=0;  v[first]:=1;  while tail<>0 do    begin      t:=ls[list[tail]];      qe:=list[tail];      tail:=tail-1;      while t>0 do        with g[t] do          begin            if d[x]+w<d[y] then              begin                d[y]:=d[x]+w;                if v[y]=0 then                  begin                    v[y]:=1;                    tail:=tail+1;                    list[tail]:=y;                    inc(sum[y]);                    if (sum[y]>=n) then                     begin                      writeln('No');                      halt;                     end;                  end;              end;            t:=next;          end;      v[qe]:=0;    end;end;procedure add(x,y,w:longint); begin  inc(o);  g[o].x:=x;  g[o].y:=y;  g[o].w:=w;  g[o].next:=ls[x];  ls[x]:=o; end;begin   read(n);   maxx:=maxlongint;  for i:=1 to n do   begin    readln(x,y,w);    add(x-1,y,-w);    if y>maxy then maxy:=y;    if x<maxx then maxx:=x;   end;  for i:=maxx to maxy do   begin    add(i,i-1,1);    add(i-1,i,0);   end;  spfa(maxx-1);  writeln(-d[maxy]);end.