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.
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.
#include <cstdio>
#include <map>
#include <iostream>
#include<cstring>
#include<bits/stdc++.h>
#define ll long long int
#define M 6
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[]={,,,,,,,,,,,,};
int dir[][]={, ,, ,-, ,,-};
int dirs[][]={, ,, ,-, ,,-, -,- ,-, ,,- ,,};
const int inf=0x3f3f3f3f;
const ll mod=1e9+;
struct node{
int w,s,id;
};
node p[];
int dp[],path[];
bool cmp1(node a,node b){
if(a.w==b.w)
return a.s>b.s;
return a.w<b.w;
}
void output(int x){
if(!x) return ;
output(path[x]);
cout<<p[x].id<<endl;
}
int main(){
ios::sync_with_stdio(false);
int ww,ss; int pos=;
while(cin>>ww>>ss){
p[pos].w=ww; p[pos].s=ss;
p[pos].id=pos;
pos++;
}
sort(p+,p+pos,cmp1);
memset(dp,,sizeof(dp));
memset(path,,sizeof(path));
int ans1=-inf;
int pp1=;
for(int i=;i<pos;i++){
int temp=,jj=;
for(int j=;j<i;j++){
if(p[i].s<p[j].s&&p[i].w>p[j].w){
if(temp<dp[j]){
temp=dp[j];
jj=j;
}
}
}
dp[i]=temp+;path[i]=jj;
if(ans1<dp[i]){
ans1=dp[i];
pp1=i;
}
}
cout<<ans1<<endl;
output(pp1);
}