HDU 1160
InputInput contains data for a bunch of mice, one mouse per line, terminated by end of file.
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.
OutputYour program should output a sequence of lines of data; the first line should contain a number n; the remaining n lines should 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.
Sample Input
6008 1300
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900
Sample Output
4
4
5
9
7
#include <bits/stdc++.h>
#define ms(a) memset(a,-1,sizeof(a))
using namespace std; const int M = 1009; int dp[M];
int pre[M]; struct node{
int w,v,id;
bool operator <(const node &n)const{
if(w>=n.w)return true;
else if (w==n.w) return v<n.v;
else return false;
}
};
node a[M]; bool cmp(node a, node b){
return a.w < b.w;
} int main()
{
int i = 0,n = 0;
ms(pre);
while(scanf("%d%d",&a[i].w,&a[i].v) != EOF)
{
a[i].id = i+1;
i++;
n++;
}
sort(a,a+n);
for(int i=0;i<n;i++)dp[i]=1;
for(int i = 0; i < n; i ++)
for(int j = 0; j < i; j++)
{
if(a[i].w < a[j].w && a[i].v > a[j].v)
{
if(dp[i] < dp[j] + 1)
dp[i] = dp[j] + 1,pre[i] = j;
}
} int ans=-1,pos = 0;
for(int i=0;i<n;i++){
if(ans<dp[i]){
ans=dp[i];
pos=i;
}
}
printf("%d\n",dp[pos]);
while(pos!=-1){
printf("%d\n",a[pos].id);
pos=pre[pos];
}
return 0;
}
HDU 1160的更多相关文章
- HDU 1160 DP最长子序列
G - FatMouse's Speed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64 ...
- 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) ...
- 怒刷DP之 HDU 1160
FatMouse's Speed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Su ...
- 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 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 解题报告
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1160 题目意思:给出一堆老鼠,假设有 n 只(输入n条信息后Ctrl+Z).每只老鼠有对应的weigh ...
- 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 ...
- FatMouse's Speed (hdu 1160)
#include <iostream> #include <cstdio> #include <cstring> #include <algori ...
- (最长上升子序列 并记录过程)FatMouse's Speed -- hdu -- 1160
http://acm.hdu.edu.cn/showproblem.php?pid=1160 FatMouse's Speed Time Limit: 2000/1000 MS (Java/Other ...
- HDU 1160(两个值的LIS,需dfs输出路径)
传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1160 FatMouse's Speed Time Limit: 2000/1000 MS (Java/ ...
随机推荐
- 当我们直接打印定义的对象的时候,隐含的是打印toString()的返回值。
以下介绍的三种方法属于Object: (1) finalize方法:当一个对象被垃圾回收的时候调用的方法. (2) toString():是利用字符串来表示对象. 当我们直接打印定义的对象的时 ...
- Zephyr学习(一)Zephyr介绍
Zephyr是一个面向物联网的嵌入式实时操作系统(RTOS),是Linux基金会旗下的一个项目,具有以下特点: 1.安全的,灵活.高可扩展性,支持多种硬件平台(ARM.ARC.X86.xtensa.n ...
- 扁平数组构建DOM树
interface IOrganizationNode { id: string; code: string; name: string; localName: string; localNameLo ...
- C++ Addon Async 异步机制
线程队列: libuv,window 可在libuv官网下载相应版本 opencv: 编译的时候opencv的位数要和 node的bit 一致 兼容electron : node-gyp rebu ...
- N - Asteroids
Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N g ...
- 松散的css
<style> <p>hello world</p> {} p { color: red; } {} <em style="padding: 4px ...
- bypass safedog upload
这里附上两个payload: Content-Disposition: form-data; name=”up_picture”; filename=”[回车]1.php” Content-Dispo ...
- Spring AOP 随记
本周经历各种面试失败后,最后一站张建飞老大的阿里,感觉有着这般年纪不该有的垃圾履历而忧伤中,不过还是要继续加油的,毕竟他说的好,都是经历,无愧初心. 所以为了更加深入理解Spring AOP我又翻起了 ...
- 用js实现二维数组的旋转
我最近因为做了几个小游戏,用到了二维数组,其中有需求将这个二维数组正翻转 90°,-90°,180°. 本人是笨人,写下了存起来. 定义的基本二位数组渲染出来是这种效果. 现在想实现的结果是下面的效果 ...
- ubuntu下安装go环境
1.官网下载go语言安装包 地址:https://studygolang.com/dl 2.服务器上安装go 将下载下来的安装包解压到/usr/local下 tar xf go1.12.1.linux ...