POJ 1201 Intervals 差分约束
链接：http://poj.org/problem?id=1201
Intervals
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 23123 | Accepted: 8728 |
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.
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
题意：
给你n个闭区间
【a，b】 及c
要求区间内的点数必须要大于等于c
点只存在于区间内。
然后问负无穷到正无穷最多有多少点。
做法：
我们从最左边到最右边。
区间内最少有c个点，说明点(a-1)到b点 最少会增加c个点。
也就是说 b-(a-1)>=c 转换下 (a-1)-b<=-c 就是差分约束的公式了。
然后就是建边 建一条权值为-c 的b到(a-1)的边就行了。
然后取最小的坐标x，取最大的坐标d。
因为求最大值 所以得公式 d-(x-1)>=?
加个负号， (x-1)-d<=-? ,所以最后最短路 计算 d点到(x-1)的最短距离，然后取负就是答案了。
但是这样条件还不够。
还有关键的是相邻点之间的建边，设相邻两点 (i) 和(i+1)
走到(i+1)数到的点数 肯定是比左边的(i)要大的 ， 所以 (i+1)-(i)>=0 加个负号， (i)-(i+1)<=0
而从(i)走到(i+1)最多只增加了一个点，所以(i+1)-(i)<=1 ，
所以还要建这些边，才能跑出最后的答案。
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