动态规划:HDU1160-FatMouse's Speed(记录动态规划状态转移过程)
FatMouse's Speed
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5546 Accepted Submission(s): 2393
Special Judge
but the speeds are decreasing.
The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information
for at most 1000 mice.
Two mice may have the same weight, the same speed, or even the same weight and speed.
m[n] then it must be the case that
W[m[1]] < W[m[2]] < ... < W[m[n]]
and
S[m[1]] > S[m[2]] > ... > S[m[n]]
In order for the answer to be correct, n should be as large as possible.
All inequalities are strict: weights must be strictly increasing, and speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one.
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900
4
5
9
7
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1010;
struct mice
{
int w,s;
int pos;
} m[maxn]; struct Dp
{
int num[maxn];
int Num;
} dp[maxn];
bool cmp(mice a,mice b)
{
return a.w<b.w;
}
int main()
{
int t = 1;
int W,S;
while(scanf("%d%d",&W,&S) != EOF)//处理这个题输入的的办法
{
m[t].w = W;
m[t].s = S;
m[t].pos = t;
t++;
} /*
可以输入0 0结束看看自己的输入是否正确
while(scanf("%d%d",&W,&S) && W)//处理这个题输入的的办法
{
m[t].w = W;
m[t].s = S;
m[t].pos = t;
t++;
}
t--;
for(int i=1;i<=t;i++)
printf("%d %d\n",m[i].w,m[i].s); */ t--;//用来记录有多少个元素
int Max = 0;//记录最长的递减子序列
sort(m,m+t,cmp);//拍一下序,cmp函数自己写一下
for(int i=1;i<=t;i++)
{
dp[i].Num = 1;
dp[i].num[1] = m[i].pos;
}
for(int i=1; i<=t; i++)
{
for(int j=0; j<=i; j++)
{
if(m[j].s > m[i].s && m[j].w < m[i].w)
{
if(dp[j].Num+1 > dp[i].Num)//如果符合状态的转移要求
{
for(int k=1;k<=dp[j].Num;k++)//记录状态的数组一起跟着转移
{
dp[i].num[k] = dp[j].num[k];
dp[i].num[k+1] = m[i].pos;//当前状态也要放入
}
dp[i].Num = dp[j].Num + 1;
if(dp[i].Num > Max)
Max = dp[i].Num;
}
}
}
} bool flag = false;
printf("%d\n",Max);
for(int i=1;i<=t;i++)
{
if(dp[i].Num == Max)
{
flag = true;
for(int j=1;j<=dp[i].Num;j++)
{
if(dp[i].num[j]!=0)
printf("%d\n",dp[i].num[j]);
}
break;
}
}
return 0;
}
动态规划:HDU1160-FatMouse's Speed(记录动态规划状态转移过程)的更多相关文章
- HDU 1160 FatMouse's Speed (动态规划、最长下降子序列)
FatMouse's Speed Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- HDU1160 FatMouse's Speed —— DP
题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=1160 FatMouse's Speed Time Limit: 2000/1000 MS ...
- HDU1160:FatMouse's Speed(最长上升子序列,不错的题)
题目:http://acm.hdu.edu.cn/showproblem.php?pid=1160 学的东西还是不深入啊,明明会最长上升子序列,可是还是没有A出这题,反而做的一点思路没有,题意就不多说 ...
- [HDU1160]FatMouse's Speed
题目大意:读入一些数(每行读入$w[i],s[i]$为一组数),要求找到一个最长的序列,使得符合$w[m[1]] < w[m[2]] < ... < w[m[n]]$且$s[m[1] ...
- 读懂TCP状态转移
读懂TCP状态转移过程,对理解网络编程颇有帮助,本文将对TCP状态转移过程进行介绍,但各状态(总共11个)含义不在本文介绍的范围,请参考文末的书目列表. TCP状态转换图(state transiti ...
- hdu FatMouse's Speed 动态规划DP
动态规划的解决方法是找到动态转移方程. 题目地址:http://acm.hdu.edu.cn/game/entry/problem/show.php?chapterid=3§ionid ...
- hdoj1160 FatMouse's Speed 动态规划
FatMouse's Speed Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- HDU 1160 FatMouse's Speed(要记录路径的二维LIS)
FatMouse's Speed Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- HDU 1160:FatMouse's Speed(LIS+记录路径)
FatMouse's Speed Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
随机推荐
- Lambda动态排序通用方法
using System; using System.Collections.Generic; using System.Linq; using System.Linq.Expressions; us ...
- iOS开发之数据存取
http://www.cocoachina.com/ios/20141111/10190.html
- VMware 虚拟机安装及部署
Linux系统安装及网络配置 这篇文章介绍关于Linux系统的安装以及网络配置,关于虚拟机配置中网络的几个模式区别进行详细讲解.学习Linux对于后端开发人员来说是很有必要的,结合实际开发,Linux ...
- es6-Iterator与for...of
Iterator(遍历器)的概念 JavaScript原有的表示“集合”的数据结构,主要是数组(Array)和对象(Object),ES6又添加了Map和Set.这样就有了四种数据集合,用户还可以组合 ...
- Spring 的AOP
AOP:面向切面编程,相对于OOP面向对象的编程 Spring的AOP的存在的目的是为了解耦.AOP可以让一组类共享相同的行为.在OOP中只能通过继承类和实现接口,来使代码的耦合度增强,且类继承只能为 ...
- LeetCode Single Number III (xor)
题意: 给一个数组,其中仅有两个元素是出现1次的,且其他元素均出现2次.求这两个特殊的元素? 思路: 跟查找单个特殊的那道题是差不多的,只是这次出现了两个特殊的.将数组扫一遍求全部元素的异或和 x,结 ...
- 从.net到java,从基础架构到解决方案。
这一年,职业生涯中的最大变化,是从.net到java的直接跨越,是从平台架构到解决方案的不断完善. 砥砺前行 初出茅庐,天下无敌.再学三年,寸步难行.很多时候不是别人太强,真的是自己太弱,却不自知. ...
- POJ 3181 Dollar Dayz(递推,两个long long)
题意:John有N美元,有价格为1~K的工具,可以买的个数不限,问1~K组合出N的方案数. f[i = 第i中工具][j = 花费为j] = 方案数. f[i][j] = sigma{ f[i-1][ ...
- 实现带复选框的TreeView控件
实现效果: 知识运用: TreeView控件的CheckView属性 //是否在树形视图控件中显示复选框 public bool CheckBoxs{ get;ser } 实现代码: TreeView ...
- 使用webpack从零开始搭建react项目
webpack中文文档 webpack的安装 yarn add webpack@3.10.1 --dev 需要处理的文件类型 webpack常用模块 webpack-dev-server yarn a ...