HDU 1160 FatMouse's Speed（DP）

题目链接：http://hdu.hustoj.com/showproblem.php?pid=1160
题目大意：输入到文件结束，每行一个老鼠的重量和速度，要求选最多只老鼠，重量依次增加，速度依次减小

[align=left]Problem Description[/align]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.
 
[align=left]Input[/align]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. 
 
[align=left]Output[/align]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
then it must be the case that 

W[m[1]] < W[m[2]] < ... < W[m
]

and 

S[m[1]] > S[m[2]] > ... > S[m
]

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. 
 
[align=left]Sample Input[/align]

6008 1300
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900 
[align=left]Sample Output[/align]

4
4
5
9
7 
[align=left]Source[/align]Zhejiang University Training Contest 2001

思路：按照速度y从大到小排个序之后，就转化成求x的最长上升子序列了，记录一下idx，最后从后往前找递归输出
代码：#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e6+5;

struct node
{
int x;
int y;
int idx;
}a[1005];

int ans[1005];
int ansy[1005];
int dp[1005];
int vis[1005];

bool cmp(node A,node B)
{
if(A.y==B.y)
return A.x<B.x;
return A.y>B.y;
}

void output(int x,int now)
{
if(x==0)
return ;
for(int i=now;i>=1;i--)
{
if(dp[i]==x)
{
output(x-1,i-1);
printf("%d\n",a[i].idx);
break;
}
}
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
std::ios::sync_with_st
4000
dio(false);
int cnt=1;
while(~scanf("%d%d",&a[cnt].x,&a[cnt].y))
{
a[cnt].idx=cnt;
cnt++;
}
cnt--;
//printf("cnt=%d\n",cnt);
sort(a+1,a+1+cnt,cmp);
for(int i=1;i<=cnt;i++)
dp[i]=1;
int ans=1;
for(int i=2;i<=cnt;i++)
{
for(int j=1;j<i;j++)
{
if(a[i].x>a[j].x&&a[i].y<a[j].y)
dp[i]=max(dp[i],dp[j]+1);
}
ans=max(ans,dp[i]);
}
printf("%d\n",ans);
int x=ans;
output(ans,cnt);
return 0;
}