（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/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12338    Accepted Submission(s): 5405
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

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>

using namespace std;
#define N 10005
#define oo 0x3f3f3f3f

/**

感觉有点用导弹防御的思想
每次记录从 0 到 i 的最长个数
并用 pre[i] 记录 i 点是从哪个点过来的

**/

/// number 是记录它原来的位置

struct node
{
    int w, s, number;
}a[N];

/// pre[] 是个很好的记录方式，可以记录它的是从哪个过来的（它的上一步）

int dp[N], pre[N];

/// 先按 w 从小到大排， 再按 s 从大到小排
int cmp(node n1, node n2)
{
    if(n1.w!=n2.w)
       return n1.w < n2.w;
    return n1.s > n2.s;
}

int main()
{
    int i=, j, n;

    memset(a, , sizeof(a));
    memset(dp, , sizeof(dp));

    while(scanf("%d%d", &a[i].w, &a[i].s)!=EOF)
    {
        a[i].number = i;
        i++;
    }

    n=i;
    sort(a+, a+i+, cmp);

/**
    for(i=1; i<n; i++)
        printf("%d %d %d\n", a[i].number, a[i].w, a[i].s);
**/

    for(i=; i<n; i++)
    {
        pre[i] = i;
        dp[i] = ;
        for(j=; j<i; j++)
        {
            if(a[i].w!=a[j].w && a[i].s<a[j].s)
            {
                if(dp[i]<dp[j]+)
                {
                     pre[i] = j;
                     dp[i] = dp[j] + ;
                }
            }
        }
    }

    int Max=-oo, index=;

    for(i=; i<n; i++)
    {
        if(dp[i]>Max)
        {
            Max = dp[i];
            index = i;
        }
    }

    printf("%d\n", Max);
    i=;
    int b[N]={};
    while(pre[index]!=index)
    {
        b[i++] = a[index].number;
        index = pre[index];
    }

    printf("%d\n", a[index].number);
    for(j=i-; j>=; j--)
        printf("%d\n", b[j]);

    return ;
}