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-6;
const int MS=50005;

struct edge
{
      int u,v,w;
}edges[10*MS];

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

bool bellman()
{
      memset(dis,0,sizeof(dis));
      bool flag=true;
      int cnt=0;
      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=0;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-1]+1<dis[i])
                  {
                        dis[i]=dis[i-1]+1;
                        flag=true;
                  }
            }

            //   i--> i-1   0

            for(int i=maxv;i>=minv;i--)
            {
                  if(dis[i]<dis[i-1])
                  {
                        dis[i-1]=dis[i];
                        flag=true;
                  }
            }
      }
      return true;
}



int main()
{
      while(scanf("%d",&n)!=EOF)
      {
            int u,v,w;
            esize=0;
            maxv=0;
            minv=INF;
            for(int i=0;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-1;
                  edges[esize++].w=-w;
                  if(v>maxv)
                        maxv=v;
                  if(u<minv)
                        minv=u;
            }
            bellman();
            printf("%d
",dis[maxv]-dis[minv-1]);
      }
      return 0;
}