http://poj.org/problem?id=2195 对km算法不理解,模板用的也不好. 下面是大神的解释. KM算法的要点是在相等子图中寻找完备匹配,其正确性的基石是:任何一个匹配的权值之和都不大于所有顶点的顶标之和,而能够取到相等的必然是最大权匹配. 左右两边点数不等时,KM算法的正确性也是可以得到保证的.原因如下: 由KM算法中可行点标的定义,有: 任意匹配的权值 <= 该匹配所覆盖的所有点的顶标值 <= KM算法所得到的匹配所覆盖的所有点的顶标值 = KM算法所得到的的匹配的权…
题面 On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves,…
http://poj.org/problem?id=2175 Evacuation Plan Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3256   Accepted: 855   Special Judge Description The City has a number of municipal buildings and a number of fallout shelters that were build…
题面 Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different k…
On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, unt…
Intervals Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 5762   Accepted: 2288 Description You are given N weighted open intervals. The ith interval covers (ai, bi) and weighs wi. Your task is to pick some of the intervals to maximize t…
题目给了一个满足最大流的残量网络,判断是否费用最小. 如果残量网络中存在费用负圈,那么不是最优,在这个圈上增广,增广1的流量就行了. 1.SPFA中某个点入队超过n次,说明存在负环,但是这个点不一定在负环上. 2.这个负环可能包括汇点t,所以构建残量网络的时候也要考虑防空洞到t上的容量. //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring…
[题目链接] http://poj.org/problem?id=3680 [题目大意] 有N个带权重的区间,现在要从中选取一些区间, 要求任意点都不被超过K个区间所覆盖,请最大化总的区间权重. [题解] 我们将权重取负后进行建图,对于每个区间从首到末连边, 如果该路被增广则说明这个区间被选定,我们只要给定K的流量 最后求出最大流下的最小费用即可 [代码] #include <cstdio> #include <algorithm> #include <cstring>…
题目链接 给一个图, N个点, m条边, 每条边有权值, 从1走到n, 然后从n走到1, 一条路不能走两次,求最短路径. 如果(u, v)之间有边, 那么加边(u, v, 1, val), (v, u, 1, val), val是路的长度,代表费用, 1是流量. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm&g…
题目描述 n个数字,求不相交的总和最大的最多k个连续子序列. 1<= k<= N<= 1000000. 输入 输出 样例输入 5 2 7 -3 4 -9 5 样例输出 13   根据贪心的思想可以知道对于一段连续的正数或负数一定是一起选或者一起不选,那么我们可以将原序列连续的正数或负数缩成一个数,并将中间的$0$及两端的负数去掉,这样序列就变成了正负正负……负正的形式.先贪心地将所有正数选取,如果正数个数$\le k$直接输出正数和就是最优方案,否则我们需要去掉一些正数或选取一些两个正数…