FatMouse's Speed Hdu1160 动态规划 EOF（Ubuntu 下codeblocks 按 Ctrl+D）

FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16982    Accepted Submission(s): 7511
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 13006000 2100500 20001000 40001100 30006000 20008000 14006000 12002000 1900

 

Sample Output

44597

 

Source

Zhejiang University Training Contest 2001

 

Recommend

Ignatius
PS：Ubuntu 下codeblocks  按 Ctrl+D 和windows下按Ctrl+Z 的作用应该是一样的，输入到文件结束。。。
tho:
按照题意先排序（体重增加，速度减小的顺序） ， 然后和最长上升子序列LIS 的做法一样即可求出答案。
重点是打印出一组解。。。
像极了当初做BFS不知道怎么存路径的时候。
翻了篇博客：http://blog.csdn.net/weizhuwyzc000/article/details/45823031
提到用紫书167页例题14的方法
发现这个和啊哈算法数组模拟邻接表的方法有些像。。。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1005;struct node{    int w,s;    int num;};bool cmp(node a, node b){    if(a.w!=b.w) return a.w<b.w;    else return a.s>b.s;}int main(){    int n;    node q[N];    int o=0;    while(scanf("%d%d",&q[o].w,&q[o].s)!=EOF)    {        q[o].num = o+1;        o++;    }    int dp[N]= {1};    int pre[N]= {0};    sort(q,q+o,cmp);    int ans=0, index = 0;    for(int i=0; i<o; i++)    {        dp[i]=1;        for(int j=0; j<i; j++)        {            if(q[i].w>q[j].w&&q[i].s<q[j].s)            {                if(dp[i]<dp[j]+1)                {                    pre[i] = j;                    dp[i] = dp[j]+1;                }            }            if(ans<dp[i])            {                ans = dp[i];                index = i;            }        }    }    cout<<ans<<endl;    int k=0;    int fina[N];    while(index != 0)    {        fina[k++] = index;        index = pre[index];    }    if(ans==1) printf("%d\n",dp[0]);    while(k>0)    {        k--;        printf("%d\n",q[fina[k]].num);    }    return 0;}