Intervals poj 1201 差分约束系统

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 22503		Accepted: 8506

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

 

设   s[i]表示集合中小于等于i的元素个数，  u   v   w      s[v]-s[u-1]>=w;

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <sstream>
 #include <iomanip>
 using namespace std;
 const int INF=0x4fffffff;
 const int EXP=1e-;
 const int MS=;

 struct edge
 {
       int u,v,w;
 }edges[*MS];

 int maxv,minv;
 int dis[MS];
 int esize,n;

 bool bellman()
 {
       memset(dis,,sizeof(dis));
       bool flag=true;
       int cnt=;
       while(flag)
       {
             flag=false;   //表示没有更新了

             if(cnt++>n)    //更新次数大于n-1,
                   return false;
             for(int i=;i<esize;i++)
             {
                   if(dis[edges[i].u]+edges[i].w<dis[edges[i].v])
                   {
                         dis[edges[i].v]=dis[edges[i].u]+edges[i].w;
                         flag=true;
                   }
             }

             //0<=s[i]-s[i-1]<=1      这些边可以不用存储

             //  i-1   -->i  1

             for(int i=minv;i<=maxv;i++)
             {
                   if(dis[i-]+<dis[i])
                   {
                         dis[i]=dis[i-]+;
                         flag=true;
                   }
             }

             //   i--> i-1   0

             for(int i=maxv;i>=minv;i--)
             {
                   if(dis[i]<dis[i-])
                   {
                         dis[i-]=dis[i];
                         flag=true;
                   }
             }
       }
       return true;
 }

 int main()
 {
       while(scanf("%d",&n)!=EOF)
       {
             int u,v,w;
             esize=;
             maxv=;
             minv=INF;
             for(int i=;i<n;i++)
             {
                   scanf("%d%d%d",&u,&v,&w);
                   //   s[i]表示集合中小于等于i的元素个数
                   //s[v]-s[u-1]>=w;    v->u-1   -w;
                   edges[esize].u=v;
                   edges[esize].v=u-;
                   edges[esize++].w=-w;
                   if(v>maxv)
                         maxv=v;
                   if(u<minv)
                         minv=u;
             }
             bellman();
             printf("%d\n",dis[maxv]-dis[minv-]);
       }
       return ;
 }