摘要: 题意比较复杂就不多说了...分析:只是拆点即可,对于@说不能同时有两个人站在上面,可以直接忽略即可,因为这个是不沉没的,故一个一个通过即可.// File Name: 11380.cpp// Author: Zlbing// Created Time: 2013/4/26 15:12:42#include<iostream>#include<string>#include<algorithm>#include<cstdlib>#include<cstdio>#include<set>#include<map># 阅读全文
posted @ 2013-04-26 16:45 z.arbitrary 阅读(483) 评论(0) 推荐(0) 编辑
摘要: 题意:N个点(1<=N<=100)M(1<=M<=5000)条有向边,运送K(1<=K<=100)个单位商品,每条边都有一个系数ai(0<ai<=100)和容量ci(ci<=5),若运送x个单位的话,就得交ai*x^2美元.问你,将K个单位商品从1运到N,最小花费.若运送不到,输出-1;分析:因为容量Ci比较小,故可以拆边,把容量为Ci的边拆成Ci条边,每条边的花费为xi*xi*ai-(xi-1)(xi-1)*ai;详细见代码// File Name: 1486.cpp// Author: Zlbing// Created Time: 20 阅读全文
posted @ 2013-04-26 14:48 z.arbitrary 阅读(221) 评论(0) 推荐(0) 编辑