(-￣▽￣)-* //求一个集合,这个集合与任意一个区间的交集,需至少有两个数字 //贪心过程:按n个区间的最右值从小到大对区间进行排列, //集合首先取第一个区间的最右两个数字, //到第二个区间,判断集合里的数有没有在区间里 //没有的话,就从第二个区间的最右开始往左取(cnt=0取最后两个数,cnt=1取最后一个数) #include<iostream> #include<cstdio> #include<cstring> #include<algorith…

Integer Intervals Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12192   Accepted: 5145 Description An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b. Write a program that: finds t…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Integer Intervals Time Limit: 1000MS   Memory Limit: 10000K Description An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b. Write a prog…

id=1716">[POJ 1716]Integer Intervals(差分约束系统) Integer Intervals Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13425   Accepted: 5703 Description An integer interval [a,b], a < b, is a set of all consecutive integers beginning with…

1716 -- Integer Intervals 跟之前个人赛的一道二分加差分约束差不多,也是求满足条件的最小值. 题意是,给出若干区间,需要找出最少的元素个数,使得每个区间至少包含两个这里的元素. 做法就是建立(b)->(a)=-2,(i)->(i+1)=1,(i+1)->(i)=0的边,然后跑一次spfa即可. 做完的时候,因为队列开太小,所以re了一次. 代码如下: #include <iostream> #include <algorithm> #inc…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12123   Accepted: 5129 Description An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b. Write a program that: finds the minimal number…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13984   Accepted: 5943 Description An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b. Write a program that: finds the minimal number…

Chiaki has n intervals and the i-th of them is [li, ri]. She wants to delete some intervals so that there does not exist three intervals a, b and c such that a intersects with b, b intersects with c and c intersects with a. Chiaki is interested in th…

题意:给出一些区间,求一个集合的长度要求每个区间里都至少有两个集合里的数. 解法:贪心或者差分约束.贪心的思路很简单,只要将区间按右边界排序,如果集合里最后两个元素都不在当前区间内,就把这个区间内的最后两个数加入集合,如果只有一个元素在区间里就加一个,如果两个元素都在区间里就不加. 差分约束系统用来解一个不等式组,只要这个不等式组里的不等式形如x1 - x2 <= c,c为常数就可以用差分约束来解不等式,将每个变量看做点,每个不等式看做边,求一个最短路,那么dis[i]就是不等式的一组解,主要利…

http://poj.org/problem?id=1716 (题目链接) 题意 给出n个区间,要求取出最少数量的不同的自然数,使每个区间中至少包含2个取出的数. Solution 差分约束. 运用前缀和,将问题转化为了一些不等式,然后建图连边跑SPFA最长路(因为是>=)即可,因为有负权所以用不了dijistra.就是poj1201的简化版. 代码 // poj1716 #include<algorithm> #include<iostream> #include<c…