HDU 1160 FatMouse's Speed (動規+最長遞減子序列)

FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9174    Accepted Submission(s): 4061
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

 

這題是一道簡單的DP題，是dp中叫最長有序子序列的問題。和數塔類似。
先將重量從小到大排，再找速度的最長遞減子序列。

代碼：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:第幾個數。
}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++;             //這道題輸入有點問題，要強制結束纔有輸出。
    }
       sort(map+1,map+i,cmp); //對數據進行排序。
       n=i-1;maxlen=0;        //n表示元素的個數。maxlen表示當前的最長的長度。
       int maxi;              //maxi記錄最長遞減子序列的末元素。
       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]) //發現以i結束的有更長的就更新。
       {
           dp[i]=dp[j]+1;
           pre[i]=j;            //pre[]記錄的就是i元素的前一個元素是j。
           if(dp[i]>maxlen)     //判斷是否比最長個還長。
           {
               maxi=i;
               maxlen=dp[i];
           }
       }
       int t=maxi;   
       i=0;
       while(t!=0)  //通過回溯將元素反序輸入res[]，同時i找到了元素的個數。
       { 
           res[i++]=t; 
           t=pre[t];
       }
       printf("%d\n",i);
       while(i>0)
       {
           i--;  //正序輸出。
           printf("%d\n",map[res[i]].index);
       }
    return 0;
}