[POJ1201 Intervals]

[题目来源]：POJ1201

[关键字]：差分约束系统

[题目大意]：有n个区间 ，已知每个区间至少有多少点问满足所有条件的最小点数。

//=====================================================================================================

[分析]：/*设s[i]为前从1到i有几个点，则可写出不等式：s[ed]-s[st-1]>=a，a就是要求的每个区间的至少有多少。同时为了将所有点连成一个图，还需要一个条件：s[i]-s[i-1]<=1。由此可以构建差分约束系统，从1开始求最短路即可。此题不可反向连边求最大路，因为：s[i]-s[i-1]<=1，如果求最大路会出现正权回路。*/以前对于差分约束系统理解不是很好，现在重新写一下。首先对于这道题可以写出不等式组：

s[i]-s[i-1]>=0 => s[i]-s[i-1]>=0

s[i]-s[i-1]<=1 => s[i-1]-s[i]>=-1

s[y]-s[x-1]>=c => s[y]-s[x-1]>=c

s[i]-s[s]>=0 => s[i]-s[s]>=0

然后连边spfa求解。具体连边可以参考http://www.cnblogs.com/procedure2012/archive/2012/03/14/2396254.html

[代码]：

type
  rec = record
    y, d, n: longint
  end;
var
  max, n, tot: longint;
  e: array[0..600000] of rec;
  link: array[0..600000] of longint;
  q: array[0..300000] of longint;
  b: array[0..600000] of boolean;
  d: array[0..600000] of longint;

procedure make(x, y, d: longint);
begin
  inc(tot);
  e[tot].y := y;
  e[tot].d := d;
  e[tot].n := link[x];
  link[x] := tot;
end;

procedure init;
var
  i, x, y, d: longint;
begin
  readln(n);
  max := 0;
  for i := 1 to n do
    begin
      readln(x,y,d);
      make(y+1,x,-d);
      if max < y+1 then max := y+1;
    end;
  for i := 0 to max do
    begin
      make(i,i+1,1);
      make(i+1,i,0);
    end;
end;

procedure spfa;
var
  h, t, x, y, next: longint;
begin
  h := 0;
  t := 1;
  fillchar(d,sizeof(d),100);
  fillchar(b,sizeof(b),false);
  b[max] := true;
  q[1] := max;
  d[max] := 0;

  while h <> t do
    begin
      h := (h+1) mod 30000;
      x := q[h];
      next := link[x];
      while next > 0 do
        begin
          y := e[next].y;
          if d[x]+e[next].d < d[y] then
            begin
              d[y] := d[x]+e[next].d;
              if not b[y] then
                begin
                  t := (t+1) mod 30000;
                  q[t] := y;
                  b[y] := true;
                end;
            end;
          next := e[next].n;
        end;
      b[x] := false;
    end;
  writeln(-d[1]);
  //readln;
end;

begin
  init;
  spfa;
end.

 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

const int MAXN=51000;

struct node
{
       int x,y,next,d;
}e[MAXN*4];
int first[MAXN];
int n,tot,m,s=51000;
int d[MAXN];
bool b[MAXN];

void Add(int x,int y,int d)
{
     e[++tot].y=y;
     e[tot].d=d;
     e[tot].next=first[x];
     first[x]=tot;
}

int SPFA()
{
    queue<int> q;
    memset(d,200,sizeof(d));
    memset(b,0,sizeof(b));
    d[s]=0,b[s]=1;
    q.push(s);
    while (!q.empty())
    {
          int u=q.front();
          b[u]=0;
          q.pop();
          for (int i=first[u];i;i=e[i].next)
          {
              int v=e[i].y;
              if (d[v]<d[u]+e[i].d)
              {
                                   d[v]=d[u]+e[i].d;
                                   if (!b[v])
                                   {
                                             q.push(v);
                                             b[v]=1;
                                   }
              }
          }
    }
    return d[n];
}

int main()
{
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    scanf("%d",&m);
    int x,y,z;
    for (int i=1;i<=m;++i)
    {
        scanf("%d%d%d",&x,&y,&z);
        Add(x-1,y,z);
        if (y>n) n=y;
        if (s>x-1) s=x-1;
    }
    for (int i=s+1;i<=n;++i)
    {
        Add(i-1,i,0);
        Add(i,i-1,-1);
        Add(s,i,0);
    }
    /*printf("%d %d\n",s,n);
    for (int i=s;i<=n;++i)
        for (int j=first[i];j;j=e[j].next)
            printf("%d %d %d\n",i,e[j].y,e[j].d);*/
    printf("%d\n",SPFA());
    return 0;
}