hdu FatMouse's Speed

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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.

题意：

给出不确定的N对MOUSE的体重和速度，要不找出一个满足体重不段上升而且速度不段下降的最大序列，并要求记录此序列号，要求输出。

分析:

在按照体重从小到大排好序之后，此问题就变成了一个求最大降序的问题，而此问题一定可以用动态规划解决。

如果我用DP【I】表示排完序列之后第I个MOUSe中所能满足的最大子降序列。因此只要在满足 list[j].wightlist[i].speed 

就可以得出状态转移方程为  DP【I】=DP【J】+1 ，并用一个数记录此条路径。

#include
#include
using namespace std;
struct node
{
    int wight;
    int speed;
    int num;
}list[1010];

bool cmp(node a,node b)
{
    if(a.wight==b.wight ) return a.speed>b.speed ;
    return a.wight < b.wight ;
}
int dp[1010],dir[1010],way[1010];

int main()
{
    int w,s;
    int i,j,k;
    int ts=1;
    int sum=-100;
    memset(dp,0,sizeof(dp));
    memset(dir,0,sizeof(dir));
    while(scanf("%d %d",&w,&s)!=EOF)
    {
        list[ts].speed = s;
        list[ts].wight = w;
        list[ts].num =ts;
        ts++;
    }
    sort(list,list+ts,cmp);
//    for(i=1;i list[i].speed && dp[j]+1>dp[i])
            {
                dir[i]=j;
                dp[i]=dp[j]+1;
            }
        }
        if(dp[i]>sum)
        {
            sum=dp[i];
            k=i;
        }
    }
    printf("%d\n",sum);
    i=1;
    while(true)
    {
        way[i++]=list[k].num;
        if(k==dir[k]) break;
        k=dir[k];
    }
    for(i--;i>0;i--)
        printf("%d\n",way[i]);
    return 0;
}