（差分约束） hdu 1384

Intervals

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3038    Accepted Submission(s): 1118

Problem 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 endpoints 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 <= 50 000) - 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 <= 50 000 and 1 <= ci <= bi - ai + 1.

Process to the end of file.

 

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

 

Author

1384

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define INF 0xfffffff
vector<int> e[50001],w[50001];
queue<int> q;
int n,dist[50001],r,l,vis[50001];
void addedge(int a,int b,int c)
{
      e[b].push_back(a);
      w[b].push_back(c);
}
void SPFA()
{
      int x;
      vis[r]=1;
      q.push(r);
      while(!q.empty())
      {
          x=q.front(),q.pop();
          vis[x]=0;
          for(int i=0;i<e[x].size();i++)
          {
                if(dist[x]+w[x][i]<dist[e[x][i]])
                {
                        dist[e[x][i]]=dist[x]+w[x][i];
                        if(!vis[e[x][i]])
                        {
                              vis[e[x][i]]=1;
                              q.push(e[x][i]);
                        }
                }
          }
      }
}
int main()
{
      while(scanf("%d",&n)!=EOF)
      {
            int a,b,c;
            l=INF,r=0;
            for(int i=0;i<50001;i++)
            {
                  vis[i]=0;
                  e[i].clear(),w[i].clear();
                  dist[i]=INF;
            }
            for(int i=0;i<n;i++)
            {
                  scanf("%d%d%d",&a,&b,&c);
                  addedge(a-1,b,-c);
                  if(a<l) l=a;
                  if(b>r) r=b;
            }
            l--;
            for(int i=l;i<r;i++)
            {
                  addedge(i+1,i,1);
                  addedge(i,i+1,0);
            }
            dist[r]=0;
            while(!q.empty()) q.pop();
            SPFA();
            printf("%d
",-dist[l]);
      }
      return 0;
}