FatMouse's Speed HDU

 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]]andS[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

题意：最长子序列题，题目给出体重和速度，所以需要先对一个量进行排序，然后剩下的那个量就可以像处理最长子序列那样做了。 值得一提的是该题需要打印路径，最好的方法是用一个数组pre记录路径。Dp是利用之前计算的结果进行递推得到的，因此，每一步的计算都要用到上一步的结。最长上升子序列，就是枚举当前序列的最后一位，然后从前面递推找最优解。

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<algorithm>using namespace std;int dp[1010],pre[1010],ans[1010];struct node{    int w,s,id;//id为初始位置，因为最后要输出，所以需要标记一下}g[1010];int cmp(node x,node y){    if(x.w != y.w)    return x.w < y.w;    else    return x.s > y.s;}int main(){   int i = 1;   while(~scanf("%d %d",&g[i].w,&g[i].s))   {       g[i].id = i;       dp[i] = 1;       pre[i] = 0;       i++;   }   int n = i-1;   sort(g+1,g+n+1,cmp);   int maxlen = 0;//最长序列长度   int maxi;//最长序列的最后一个数下标   dp[1] = 1;   for(i=1;i<=n;i++)   {       for(int j=1;j<i;j++)       {            if(g[i].w>g[j].w&&g[i].s<g[j].s&&dp[i]<dp[j]+1)            {                dp[i] = dp[j]+1;                pre[i] = j;                if(dp[i]>maxlen)                {                     maxlen = dp[i];                     maxi = i;                }            }       }   }   int t = maxi;   i=0;   while(t!=0)   {       ans[i++] = t;       t = pre[t];   }   printf("%d\n",i);   while(i>0)   {       i--;       printf("%d\n",g[ans[i]].id);   }   return 0;}