fir pre ive art name namespace 體重 oss ()

傳送門：

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): 20100 Accepted Submission(s): 8909
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

Source Zhejiang University Training Contest 2001 題目意思： 找到一個最多的老鼠序列，使得序列中的老鼠的體重滿足遞增，相應老鼠的速度滿足遞 減。即可要求找出老鼠體重遞增,速度遞減的最長子序列(不需要連續). 分析： 最大上升子序列，先按Wi sort一下，然後LIS，最後dfs輸出該序列
code：

#include<bits/stdc++.h>
using namespace std;
#define max_v 10050
struct node
{
    int w,s,index;
}m[max_v];
int pre[max_v];
int dp[max_v];
bool cmp(node a,node b)
{
    if(a.w!=b.w)
        return a.w<b.w;
    else
        return a.s<b.s;
}
void dfs(int i)
{
    int num=m[i].index;
    if(i!=pre[i])
    {
        dfs(pre[i]);
    }
    printf("%d\n",num);
}
int main()
{
    //最大上升子序列，先按Wi sort一下，然後LIS，最後dfs輸出該序列
    int n=1;
    while(~scanf("%d %d",&m[n].w,&m[n].s))
    {
        m[n].index=n;
        n++;
    }
    sort(m+1,m+1+n,cmp);
    pre[1]=1;
    dp[1]=1;
    for(int i=2;i<=n;i++)
    {
        int maxx=0;
        int maxi=i;
        for(int j=i-1;j>=1;j--)
        {
            if(m[i].s<m[j].s)
            {
                if(dp[j]>maxx)
                {
                    maxx=dp[j];
                    maxi=j;
                }
            }
        }
        dp[i]=maxx+1;
        pre[i]=maxi;
    }
    int maxx=0;
    int maxi;
    for(int i=1;i<=n;i++)
    {
        if(maxx<dp[i])
        {
            maxx=dp[i];
            maxi=i;
        }
    }
    printf("%d\n",maxx);
    dfs(maxi);
    return 0;
}

HDU 1160（兩個值的LIS，需dfs輸出路徑）