HDU 1384 Intervals 差分约束

Intervals

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2139    Accepted Submission(s): 760

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

 代码 ：

// 差分约束
#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std ;
typedef pair< int , int >pii ;
#define MAX 50010
vector<pii>qe[MAX];
#define INF 25797524
int d[50010] ;
bool vi[MAX] ;
//d[t+1]-d[s] >= c,也就是dist[t+1] >= dist[s] + c,
//这是求最长路的形式,所以要求最长路 // 看形势
// 如果转化的是 d[s] - d[e] < - w ; 那么 的d[e] > d[s] + w ;
// 这求的就是最短路
// 网上看的，用SPFA解差分方程，用最短路径求差分方程的最大解；用最长路径求差分方程的最小解.
void spfa( int Max , int Min )
{
	int i , j , now  ;
	int u , w ;
	memset( vi , 0 , sizeof(vi) ) ;
	queue<int>q ;
   for( i = Min ; i <= Max ;i++ )
	    d[i] = -INF ;// 因为求最长路 
        d[Min] = 0 ;
		q.push( Min ) ;
		vi[1] = Min ;
		while( !q.empty())
		{
			now = q.front( ) ;
			q.pop( ) ;
			for( i = 0 ; i < qe[now].size() ; i++)
			{
				pii a = qe[now][i] ;
				u = a.first ;
				w = a.second ;
				if( d[u] < w + d[now] ) // 最长路形势 
				{
					d[u] = w + d[now] ;
					if(!vi[u])
					{
						q.push( u ) ;
						vi[u]= 1 ;
					}
				}
			}
			vi[now] = 0 ;
		}
}
int main()
{
	int i , j , n ,m ;
	int u , v , w , len,  Max = MAX , Min ;
	while( scanf( "%d" , &n ) != EOF )
	{ 
		for( i = 0 ; i <= Max ; i++)
			qe[i].clear() ;
		Max = 0 ;Min = INF ;
		for( i = 1 ; i <= n ;i++ )
		{  			
				scanf( "%d%d%d" , &u , &v , &w ) ;
					qe[u].push_back( pii( v+1 , w ) ) ;
					if( v + 1 > Max ) 
						Max = v + 1 ;
					if( u < Min ) Min = u ;

		}
		for( i = Min ; i < Max ; i++)
		{  
			qe[i+1].push_back(pii( i , -1 )) ;
			qe[i].push_back( pii( i + 1 , 0 ) ) ;
		}
	   spfa( Max , Min ) ;
	   printf( "%d\n" ,d[Max] - d[Min] ) ;
	} 
}