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 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.
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.
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7解題思路：首先根據weight的數值由小到大排序，然后再以speed為標準，求最長下降子序列，思路還是DP，但是要標記子序列的下標麻煩多了，我的做法：定義MaxIncludeEnd數組，令MaxIncludeEnd[i]表示以a[i]結尾的最長下降子序列的長度，那么MaxIncludeEnd[i]=max{ MaxIncludeEnd[i] , a[j]>a[i] ? (MaxIncludeEnd[j]+1) ：-1; }其中j=0:i-1;MaxIncludeEnd[i]的初始值為1,每一個i結束后用max和MaxIncludeEnd[i]比較，更新max和下標max_flg。因為要標記下標，在計算MaxIncludeEnd[i]后，用flg[i]表示以a[i]結尾的最長下降子序列的前一個元素的下標，這樣計算完MaxIncludeEnd[n]并更新完max和max_flg以后，循環遞推一次就可以得到所有的元素下標，即flg[max_flg],flg[ flg[max_flg] ],……當然這里的下標是逆向的，要設置個新的數組來將其逆轉。ps：如果從后往前計算最長上升子序列，那就不必如此麻煩，減少空間。
code
#include<cstdio>
#include<algorithm>
using namespace std;
struct mouse
{
int weight;
int speed;
int num;
}mice[1002];
int MaxIncludeEnd[1002],flg[1002],rst[1002];
int cmp(struct mouse a,struct mouse b)
{
if(a.weight<b.weight)return 1;
return 0;
}
int main()
{
int w,s,i=0,j,k,size,max,max_flg;
while(scanf("%d%d",&w,&s)!=EOF)
{
mice[i].weight=w;
mice[i].speed=s;
mice[i].num=i+1;
i++;
}
size=i;
sort(mice,mice+size,cmp);
max=max_flg=flg[0]=0;
for(i=0;i<size;i++)
{
MaxIncludeEnd[i]=1;
for(j=0;j<i;j++)
{
if(mice[i].speed<mice[j].speed && mice[i].weight>mice[j].weight && MaxIncludeEnd[j]+1>MaxIncludeEnd[i])
{
MaxIncludeEnd[i]=MaxIncludeEnd[j]+1;
flg[i]=j;
}
}
if(max<MaxIncludeEnd[i])
{
max=MaxIncludeEnd[i];
max_flg=i;
}
}
printf("%d\n",max);
k=0;
while(max--)
{
rst[k++]=max_flg;
max_flg=flg[max_flg];
}
for(k--;k>=0;k--)
printf("%d\n",mice[rst[k]].num);
return 0;
}