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  • poj 1201

    差分约数

    Intervals
    差分约束详解 转 https://blog.csdn.net/whereisherofrom/article/details/78922648
    他这个一看就懂
    Intervals
    Time Limit: 2000MS Memory Limit: 65536K
    Total Submissions: 32784 Accepted: 12667
    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

                                          间隔
                           时间限制: 2000MS		内存限制: 65536K
    描述
    
    给定n个闭合的整数区间[ai,bi]和n个整数c1,...,cn。 
    编写一个程序: 
    从标准输入读取间隔数,它们的终点和整数c1,...,cn, 
    计算整数集合Z的最小尺寸,它至少具有ci公共元素的间隔[ai, bi],对于每个i = 1,2...,n, 
    将答案写入标准输出。 
    输入
    
    输入的第一行包含整数n(1 <= n <= 50000- 间隔数。以下n行描述了间隔。输入的第(i + 1)行包含由单个空格分隔的三个整数ai,bi和ci,并且使得0 <= ai <= bi <= 50000并且1 <= ci <= bi-ai + 1。
    产量
    
    对于每个i = 1,2...,n,输出恰好包含一个等于集合Z的最小尺寸的整数,该集合Z至少与区间[ai,bi]共享ci元素。
    样本输入
    
    五
    3 7 3
    8 10 3
    6 8 1
    1 3 1
    10 11 1
    样本输出
    
    6
    资源
    
    2002年西南欧
    
    
    > s(b) - s(a - 1) >= c  所以 * (-1)得 s(a - 1) - s(b) <= -c 这是条件之一   w(b , a - 1) = -c
    > 根据题目s(i) - s(i - 1) >= 0  所以 s(i - 1) - s(i) <= 0  条件二 w(i , i - 1) = 0
    > 每个位置上至多有一个元素  s(i ) -  s(i - 1) <= 1  条件三 w(i - 1 , i ) = 1
    > 然后题目的意思就是让求s(to) - s(from - 1) >= w , 最少就是w  , 乘以-1 s(from - 1) - s(to) <= -w 
    > 为了避免from - 1 < 0 所以在建图的时候全部 a++ , b ++ 也就是将所有的区间向右挪一位
    
    #include <iostream>
    #include <queue>
    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    //#pragma GCC optimize(3 , "Ofast" , "inline")
    using namespace std;
    const int N = 1e6 + 10 ;
    int e[N * 2] , ne[N * 2] , h[N * 2] , idx , w[N * 2] , dis[N] , vis[N] , n , out[N];
    bool st[N] ;
    
    void add(int a , int b , int c)
    {
    	a ++ , b ++ ;
    	e[idx] = b , w[idx] = c , ne[idx] = h[a] , h[a] = idx ++ ;
    }
    queue<int> q ;
    void SPFA(int from)
    {
    	for(int i = 0;i <= N;i ++) dis[i] = 0x3f3f3f3f ;
    	q.push(from) ;
    	memset(st , false , sizeof st) ;
    	dis[from] = 0 ; 
    	st[from] = true ;
    	int flag = 0 ;
    	while(q.size())
    	{
    		int t = q.front() ;
    		q.pop() ;
            st[t] = false ;
    		for(int i = h[t] ;i != -1 ;i = ne[i])
    		{
    	//		cout << " -1 " << endl ;
    			int j = e[i] ;
    			if(dis[j] > dis[t] + w[i])
    			{
    				dis[j] = dis[t] + w[i] ;
    				if(!st[j])
    				 {
    				 	st[j] = true ;
    				 	q.push(j) ;
    				 }
    		
    			}
    		}
    		
    	}
        return ;
    }
    int main()
    {
    	int  n1 , n2 ;
    	while(~scanf("%d",&n))
    	{
    	idx = 0 ;
    	int a ,b , c ;
    	int from = 0x3f3f3f3f , to = 0 ;
    	memset(h , -1 , sizeof h);
    	for(int i = 1;i <= n ;i ++)
    	{
    		scanf("%d%d%d",&a, &b, &c) ;
    		add(b , a - 1, -c) ;
    		if(from > a) from = a ;
    		if(b > to) to = b ;
    	}
    //	cout << from << " " << to << endl ;
    	for(int i = from ;i <= to ;i ++)
    	 add(i - 1 , i , 1) , add(i , i - 1 , 0) ;
    	SPFA(to + 1) ;
    	printf("%d\n" , -dis[from]);
    	}
    	
    	return  0;
    } 
    
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  • 原文地址:https://www.cnblogs.com/spnooyseed/p/12870929.html
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