FatMouse‘s Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Special Judge

Problem Description

FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection 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.

Input

Input 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.

Output

Your 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

这题是一道简单的DP题，是dp中叫最长有序子序列的问题。和数塔类似。

先将重量从小到大排，再找速度的最长递减子序列。

代码：0MS

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 2050
struct node {
int w,s,index;  //w:重量，s:速度，index:第几个数。
}map[M];
bool cmp(node x,node y){   //如果重量相等，速度大的排前面。
 if(x.w==y.w)
   return x.s>y.s;
   return x.w<y.w;      //如果重量不相等，重量小的排前面。
}
int res[M],dp[M],pre[M];
int main()
{
    int i=1,j,k,n,m,maxlen;
    while(cin>>n>>m)
    {
       map[i].w=n;map[i].s=m;map[i].index=i;
       dp[i]=1;pre[i]=0;
       i++;             //这道题输入有点问题，要强制结束才有输出。
    }
       sort(map+1,map+i,cmp); //对数据进行排序。
       n=i-1;maxlen=0;        //n表示元素的个数。maxlen表示当前的最长的长度。
       int maxi;              //maxi记录最长递减子序列的末元素。
       dp[1]=1;
       for(i=1;i<=n;i++)      //模板。
       for(j=1;j<i;j++)
       if(map[i].w>map[j].w && map[i].s<map[j].s && dp[j]+1>dp[i]) //发现以i结束的有更长的就更新。
       {
           dp[i]=dp[j]+1;
           pre[i]=j;            //pre[]记录的就是i元素的前一个元素是j。
           if(dp[i]>maxlen)     //判断是否比最长个还长。
           {
               maxi=i;
               maxlen=dp[i];
           }
       }
       int t=maxi;
       i=0;
       while(t!=0)  //通过回溯将元素反序输入res[]，同时i找到了元素的个数。
       { 
           res[i++]=t; 
           t=pre[t];
       }
       printf("%d\n",i);
       while(i>0)
       {
           i--;  //正序输出。
           printf("%d\n",map[res[i]].index);
       }
    return 0;
}

HDU 1160 FatMouse's Speed (动规+最长递减子序列),布布扣,bubuko.com

HDU 1160 FatMouse's Speed (动规+最长递减子序列)