FatMouse's Speed

                                                   FatMouse's Speed

原题链接  http://acm.hdu.edu.cn/showproblem.php?pid=1160

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

找出矩形的左边界 l 与右边界 r ，并将左边界定义为左边连续的高度>=其最左边的矩形的下标，从左往右推出所有的左边界，同理从右往左推出所有的右边界。矩形的高度可以等于零，所以一定要在循环条件中加上限制边界的条件来终结循环，否则当左边边上的矩形高为零或者右边边界上矩形高度为零时会进入死循环。

以下是AC代码

#include
#include
#include
using namespace std;

long long  a[100010],l[100010],r[100010];

int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        int i,j;
        long long  Max=-1,sum;
        for(i=1; i<=n; i++)
            scanf("%lld",&a[i]);
        l[1]=1;
        r[n]=n;
        for(i=2; i<=n; i++)
        {
            int tt=i;
            while(tt>1&&a[i]<=a[tt-1])
                tt=l[tt-1];
            l[i]=tt;
        }
        for(i=n-1; i>=1; i--)
        {
            int tt=i;
            while(ttMax)
                Max=sum;
        }
        printf("%lld\n",Max);
    }
    return 0;
}