There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of gas
to travel from station i to ...
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其他好文 时间:
2014-07-10 20:59:16
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239
题目
There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of gas
to travel from s...
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其他好文 时间:
2014-07-01 11:02:24
阅读次数:
158
题目
There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of gas
to travel from s...
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其他好文 时间:
2014-06-30 16:57:25
阅读次数:
231
【题目】
There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an ...
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其他好文 时间:
2014-06-26 13:27:05
阅读次数:
231
把N个点先转化为2*N-2个点。
比如说把012345转化成0123454321。
这样,就可以找出任意两两个点之间的关系。
然后根据关系可以得出来一个一元多项式的矩阵。
然后就用高斯消元求出矩阵即可。
#include
#include
#include
#include
#include
#include
using namespace std;
#define eps 1e-6
#...
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其他好文 时间:
2014-06-25 07:08:28
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197
USE [Travel]GO/****** Object: StoredProcedure [dbo].[NoticeGetPagedData] Script Date: 06/13/2014 20:44:51 ******/SET ANSI_NULLS ONGOSET QUOTED_IDENTIF...
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数据库 时间:
2014-06-14 19:45:10
阅读次数:
255
The design of travel bags and promotional gifts
are supported appropriately. This means that the authorities believe that many
things in addition to d...
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其他好文 时间:
2014-05-22 02:12:01
阅读次数:
324
很多天没更新微博了,最近工作比较忙在支持一个UE3的项目,EasyKit架构也在无缝地图的方面设计更多解决方案(UnrealEngine4中自带一个seamless
travel,主要涉及到广播和LevelStreaming)。昨天查资料发现好多人在问如何编译客户端。好吧,如果不了的确要花点时间研究...
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其他好文 时间:
2014-05-15 21:29:04
阅读次数:
308
给n个点,m条边的无向图,一个起点和一个终点。每边都有消耗,经过就要付出代价;每个点有消耗和价值,只有消耗了才会获得价值,如果不消耗就不会获得价值,且下一次消耗的点的价值一定要严格大于之前消耗过的点的价值
求:起点到终点消耗不超过给定值T时的价值最大值
1 < N < 100,0 < M < 1000,0 < T <= 300...
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其他好文 时间:
2014-05-08 17:22:31
阅读次数:
332
HDU 4571
大概题意:n个点(
解法:
容易看出应该用spfa和dp来解。关键时对visit和pass点的处理。
通过floyd预处理出visit每个点对之间的最小边消耗。然后,加一个超级源点和一个超级终点。超级源点负责pas点s能够到达的点,超级终点负责那些能越过e的点
由于visit的点的moneyp值必须严格升序所以也可以拓扑之后dp
不能用dij,因为本题时求最长...
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其他好文 时间:
2014-05-08 17:12:27
阅读次数:
408