hdu 1160 FatMouse's Speed（动态规划）

问题描述：

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.

输入：

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. 

输出：

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. 

分析：

这是很典型的一道dp题，思路是先对体重进行升序排列，当体重相等时对速度进行降序排列，对速度便是利用求最大递减子序列！最大递减子序列的思路见此博客原创最大递增子序列那篇！

代码：

#include<iostream>#include<algorithm>#include<stdio.h>using namespace std;int l[1001];int flag[1001][1001];//当把两个数组放到这里时不会出现栈溢出struct Mice{    int w;    int s;    int id;};Mice mices[1001];int cmp(struct Mice a,struct Mice b){    if(a.w!=b.w) return a.w<b.w;    else return a.s>b.s;}int main(){    //int l[1001];//存储最大递减序列的长度的值    //int flag[1001][1001];//存储递减序列下表的值    int i,size,j,k,max,t;    i=0;    while(scanf("%d %d",&mices[i].w,&mices[i].s)!=EOF)    {        mices[i].id=i+1;        i++;    }    size=i;    sort(mices,mices+size,cmp);    for(i=0;i<size;i++)    {        l[i]=1;        flag[i][0]=i;    }    for(i=1;i<size;i++)    {        for(j=i-1;j>=0;j--)        {            if(mices[i].w>mices[j].w && mices[i].s<mices[j].s && l[i]<l[j]+1)            {                max=l[j]+1;                l[i]=max;                        for(k=0;k<max-1;k++)                flag[i][k]=flag[j][k];            flag[i][max-1]=i;            }        }    }    int index;        for(index=0,k=1;k<size;k++)         if(l[index]<l[k]) index=k;        cout<<l[index];        for(i=0;i<l[index];i++)        cout<<endl<<mices[flag[index][i]].id;    return 0;}

注意：除了将数组放到函数外才可以避免栈溢出外，还有一个问题是关于文件末尾的说法，该题while小括号里若是改成cin也是可以的