HDU 1160 FatMouse's Speed(DP)
题意 输入n个老鼠的体重和速度 从里面找出最长的序列 是的重量递增时速度递减
简单的DP 令d[i]表示以第i个老鼠为所求序列最后一个时序列的长度 对与每一个老鼠i 遍历全部老鼠j 当(w[i] > w[j]) && (s[i] < s[j])时 有d[i]=max(d[i],d[j]+1) 输出路径记下最后一个递归即可了
#include<cstdio>
#include<algorithm>
using namespace std;
const int M=1005;
int w[M], s[M], d[M], pre[M], n, key; int dp (int i)
{
if (d[i] > 0) return d[i];
for (int j = d[i] = 1; j <= n; ++j)
if ( (w[i] > w[j]) && (s[i] < s[j]) && (d[i] < dp (j) + 1))
d[i] = d[j] + 1, pre[i] = j;
return d[i];
} void print (int i)
{
if (pre[i])
print (pre[i]);
printf ("%d\n", i);
} int main()
{
n = 0;
while (scanf ("%d %d", &w[n], &s[++n]) != EOF);
for (int i = key =1; i <= n; ++i)
if (dp (i) > dp (key)) key = i;
printf ("%d\n", d[key]);
print (key);
return 0;
}
FatMouse's Speed
of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, 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.
each contain a single positive integer (each one representing a mouse). If these n integers are m[1], m[2],..., 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.
6008 1300
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900
4
4
5
9
7
HDU 1160 FatMouse's Speed(DP)的更多相关文章
- HDU 1160 FatMouse's Speed DP题解
本题就先排序老鼠的重量,然后查找老鼠的速度的最长递增子序列,只是由于须要按原来的标号输出,故此须要使用struct把三个信息打包起来. 查找最长递增子序列使用动态规划法.主要的一维动态规划法了. 记录 ...
- HDU 1160 FatMouse's Speed (最长有序的上升子序列)
题意:给你一系列个w,s.要你找到最长的n使得 W[m[1]] < W[m[2]] < ... < W[m[n]] and S[m[1]] > S[m[2]] > ... ...
- HDU 1160 FatMouse's Speed (DP)
FatMouse's Speed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Su ...
- HDU 1160 FatMouse's Speed(DP)
点我看题目 题意 :给你好多只老鼠的体重和速度,第 i 行代表着第 i 个位置上的老鼠,让你找出体重越大速度越慢的老鼠,先输出个数,再输出位置. 思路 :看题的时候竟然脑子抽风了,看了好久愣是没明白题 ...
- 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 动态规划LIS,路径还原与nlogn优化
HDU - 1160 给一些老鼠的体重和速度 要求对老鼠进行重排列,并找出一个最长的子序列,体重严格递增,速度严格递减 并输出一种方案 原题等于定义一个偏序关系 $(a,b)<(c.d)$ 当且 ...
- HDU 1160 FatMouse's Speed (最长上升子序列)
题目链接 题意:n个老鼠有各自的重量和速度,要求输出最长的重量依次严格递增,速度依次严格递减的序列,n最多1000,重量速度1-10000. 题解:按照重量递增排序,找出最长的速度下降子序列,记录序列 ...
- HDU 1160 FatMouse's Speed LIS DP
http://acm.hdu.edu.cn/showproblem.php?pid=1160 同样是先按它的体重由小到大排,相同就按speed排就行. 这样做的好处是,能用O(n^2)枚举,因为前面的 ...
- HDU 1160 FatMouse's Speed (sort + dp)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1160 给你一些老鼠的体重和速度,问你最多需要几只可以证明体重越重速度越慢,并输出任意一组答案. 结构体 ...
随机推荐
- python数组中数据位置交换 -- IndexError: list assignment index out of range
代码: t = [-10,-3,-100,-1000,-239,1] # 交换 -10和1的位置 t[5], t[t[5]-1] = t[t[5]-1], t[5] 报错: IndexError: l ...
- html前端如何将一个页面表单内的数据全部传递到另一个页面?
http://blog.csdn.net/stone_tomcate/article/details/64148648?winzoom=1
- 深入Linux内核架构——进程虚拟内存
逆向映射(reverse mapping)技术有助于从虚拟内存页跟踪到对应的物理内存页: 缺页处理(page fault handling)允许从块设备按需读取数据填充虚拟地址空间. 一.简介 用户虚 ...
- Python中比元组更好用的namedtuple
一.思考 1.什么是元组? 不可变的序列类型 "不能修改的列表" 2.元组支持哪些操作? 元组是序列类型,支持序列类型的所有操作 通过索引取值 one_tuple = (" ...
- viva correction statements
* List of amendments| No. | Location | Amendments ...
- 大数据学习——hdfs集群启动
第一种方式: 1 格式化namecode(是对namecode进行格式化) hdfs namenode -format(或者是hadoop namenode -format) 进入 cd /root/ ...
- FZU-2147-2147 A-B Game,规律题。。
Problem 2147 A-B Game Time Limit: 1000 mSec Memory Limit : 32768 KB Problem Description Fat brother ...
- Go变量定义学习
package main import ( "fmt" ) //变量定义: //使用var关键字或:=定义变量 //可放在函数内,或直接放在包内 //使用var()集中定义 //函 ...
- 【multimap的应用】D. Array Division
http://codeforces.com/contest/808/problem/D #include<iostream> #include<cstdio> #include ...
- mappedBy的具体使用及其含义
mappedBy: 1>只有OneToOne,OneToMany,ManyToMany上才有mappedBy属性,ManyToOne不存在该属性: 2>mappedBy标签一定是定义在被拥 ...