HDU1160 FatMouse’s Speed

                                             
FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7133    Accepted Submission(s): 3139
Special Judge

Problem DescriptionFatMouse 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. 

InputInput 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. 

OutputYour 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 13006000 2100500 20001000 40001100 30006000 20008000 14006000 12002000 1900

 

Sample Output

44597

 

SourceZhejiang University Training Contest 2001 

RecommendIgnatius

解題思路：在體重嚴格遞增的情況下，要求速度嚴格遞減的老鼠有幾隻，還要輸出他們出現的順序（按輸入時順序值）。
                       在體重與速度遞減排序後，加條件求最大嚴格遞增子序列，記錄路徑，後面直接輸出。
                     求最大子序列（http://blog.csdn.net/lsh670660992/article/details/9498833）

#include<cstdio>#include<algorithm>using namespace std;struct node    //老鼠的屬性{    int w;    //體重    int s;     //速度    int x;     //輸入時的順序號}mice[1005];int cmp(node a,node b)   //以體重為標準，將老鼠按體重遞減排序{    return a.w<b.w;}int main(){    int n=0;    int i,j;    int much[1005],head[1005];    //分別記錄第i只老鼠為止，子序列長度與前驅儲存位置（mice數組中下標）    while(scanf("%d%d",&mice[n].w,&mice[n].s)!=EOF)        mice[n].x=n+1,n++;    sort(mice,mice+n+1,cmp);   //排序    head[0]=-1;   //終止標記    much[0]=1;    int max1=0,x;    //由第一隻老鼠到當前檢索位置為止，子序列的最大長度和其前驅位置在前驅記錄數組head中的下標    for(i=1;i<=n;i++)    {        head[i]=-1;        much[i]=1;        for(j=0;j<i;j++)        {            if(mice[i].w>mice[j].w&&mice[i].s<mice[j].s&&much[j]+1>much[i])    //體重嚴格遞增，速度嚴格遞減，子序列更長            {                much[i]=much[j]+1;                head[i]=j;                if(much[i]>max1)   //前驅處理                {                    max1=much[i];                    x=i;                }            }        }    }    int road[1005];    i=0;    while(1)   //處理路勁記錄    {        road[i++]=mice[x].x;        x=head[x];        if(x==-1)            break;    }    printf("%d\n",max1);    for(i--;i>=0;i--)        printf("%d\n",road[i]);    return 0;}