poj1716 Integer intervals (differential constraint)

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Integer intervals

Time Limit: 1000MS

Memory Limit: 10000K

Description

An integer interval [A, a,], a < B, was a set of all consecutive integers beginning with a and ending with B.
Write a program that:finds the minimal number of elements in a set containing at least II different integers from each i Nterval.

Input

The first line of the input contains the number of intervals N, 1 <= n <= 10000. Each of the following n lines contains and integers a, b separated by a single space, 0 <= a < b <= 10000. They is the beginning and the end of an interval.

Output

Output the minimal number of elements in a set containing at least, different integers from each interval.

Sample Input

43 62 40) 24 7

Sample Output

4

The castrated version of the previous differential constraint (portal)

1#include <iostream>2#include <cstring>3#include <cstdio>4#include <cstdlib>5#include <algorithm>6#include <queue>7 using namespacestd;8typedefLong Longll;9typedef pair<int,int>PII;Ten #defineMAXN 50010 One #defineREP (a,x) for (int a=0; a<x; a++) A #defineINF 0x7fffffff - #defineCLR (a,x) memset (a,x,sizeof (A)) - structNode { the     intV,d,next; -}edge[3*MAXN]; - intHEAD[MAXN]; - intE=0; + voidInit () - { +REP (I,MAXN) head[i]=-1; A } at voidAdd_edge (intUintVintd) - { -edge[e].v=v; -Edge[e].d=D; -edge[e].next=Head[u]; -head[u]=e; ine++; - } to BOOLVIS[MAXN]; + intDIS[MAXN]; - voidSPFA (ints) { theCLR (Vis,0); *REP (I,MAXN) dis[i]=i==s?0: INF; $queue<int>Q;Panax Notoginseng Q.push (s); -vis[s]=1; the      while(!q.empty ()) +     { A         intx=Q.front (); the Q.pop (); +         intt=Head[x]; -          while(t!=-1) $         { $             inty=edge[t].v; -             intD=edge[t].d; -t=Edge[t].next; the             if(dis[y]>dis[x]+d) -             {Wuyidis[y]=dis[x]+D; the                 if(Vis[y])Continue; -vis[y]=1; Wu Q.push (y); -             } About         } $vis[x]=0; -     } - } - intMain () A { +Ios::sync_with_stdio (false); the     intN; -     intu,v,d; $     intans=0; the     intmaxn=0; the     intminn=MAXN; the      while(SCANF ("%d", &n)! = eof&&N) the     { -E=0; in init (); the REP (i,n) the         { Aboutscanf"%d%d",&u,&v); theAdd_edge (u,v+1,-2); theMaxn=max (maxn,v+1); theminn=min (u,minn); +         } -          for(inti=minn;i<maxn;i++){ theAdd_edge (i+1I1);BayiAdd_edge (i,i+1,0); the         } the SPFA (Minn); -cout<<-dis[maxn]<<Endl; -     } the     return 0; the}

code June

poj1716 Integer intervals (differential constraint)