动态规划(DP),类似LIS,FatMouse's Speed

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1108

解题报告：

1、首先按照weight从小到大排列，weight相同的按照speed从大到小排列;

2、Count[i]表示到第i个老鼠时，所求的最长“速度递减”子序列的长度；

3、path[i]=j是题目的关键，记录在Count[i]=Count[j]时，即最长“速度递减”子序列最后一个老鼠的前一只老鼠的位置

4、递归输出id

void output(int path[],pos)
{
    if(pos==0) return ;
    output(path,path[pos]);
    printf("%d\n",mice[pos].id);
}

Sources Code:

#include <cstdio>
#include <stdlib.h>
#include <algorithm>
#define INF 0x3f3f3f3f

using namespace std;

int n;///n只老鼠
int Count[1001]= {0}; ///count[i]表示构造到第i个老鼠时，序列的最大长度
int path[1001]= {0}; ///path[i]表示构造到第i个老鼠时的前一个老鼠(前驱)

struct mouse
{
    int weight;
    int speed;
    int id;
} mice[1001];

int cmp(const void *a,const void *b)
{
    struct mouse *p1=(mouse *)a;
    struct mouse *p2=(mouse *)b;
    if(p1->weight==p2->weight)
    {
        if(p1->speed>p2->speed)
            return p2->speed-p1->speed;
        else return p1->speed-p2->speed;
    }
    else return p1->weight-p2->weight;
}

void output(int path[],int pos)
{
    if(pos==0) return ;
    output(path,path[pos]);
    printf("%d\n",mice[pos].id);
}

int main()
{
    n=0;
    int i=0,j=0;
    while(scanf("%d%d",&mice[++i].weight,&mice[++j].speed)!=EOF)
    {
        n++;
        mice[n].id=n;
    }
    qsort(mice+1,n,sizeof(mice[0]),cmp);
    Count[1]=1;
    for(int i=2; i<=n; i++)
    {
        for(int j=1; j<i; j++)
        {
            if(mice[i].weight>mice[j].weight&&mice[i].speed<mice[j].speed)
            {
                if(Count[i]<Count[j])
                {
                    Count[i]=Count[j];
                    path[i]=j;
                }
            }
        }
        Count[i]++;
    }
    int _max=0;
    int pos;
    for(int i=1; i<=n; i++)
    {
        if(Count[i]>_max)
        {
            _max=Count[i];
            pos=i;
        }
    }
    printf("%d\n",_max);
    output(path,pos);
}

动态规划(DP),类似LIS,FatMouse's Speed