UVA 10131

Question 1: Is Bigger Smarter?

The Problem

Some people think that the bigger an elephant is, the smarter it is. To disprove this, you want to take the data on a collection of elephants and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the IQ's are decreasing.

The input will consist of data for a bunch of elephants, one elephant per line, terminated by the end-of-file. The data for a particular elephant will consist of a pair of integers: the first representing its size in kilograms and the second representing its IQ in hundredths of IQ points. Both integers are between 1 and 10000. The data will contain information for at most 1000 elephants. Two elephants may have the same weight, the same IQ, or even the same weight and IQ.

Say that the numbers on the i-th data line are W[i] and S[i]. 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 an elephant). If these n integers are a[1], a[2],..., a[n]then it must be the case that

   W[a[1]] < W[a[2]] < ... < W[a[n]]

and

   S[a[1]] > S[a[2]] > ... > S[a[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 IQs must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one.

Sample Input

6008 1300
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900

Sample Output

4
4
5
9
7

直接按一种元素排序，然后求另一个元素的LDS或LIS

可能也是赶上今天心情不好，这题居然9WA1CE， 哎 ， 我这是要爆发的节奏啊！

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <iomanip>

using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1|1
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
#define mid ((l + r) >> 1)
#define mk make_pair
const int MAXN = 1000 + 50;
const int maxw = 100 + 20;
const int MAXNNODE = 10000 +10;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
#define mod 10007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pii;
const D e = 2.718281828459;
int dp[MAXN];
int path[MAXN];
struct elephant
{
    int w,s , num;
}a[MAXN];
int cmp (const elephant a , const elephant b)
{
    if(a.w != b.w)return a.w < b.w;
    return a.s > b.s;
}

int main()
{
    //ios::sync_with_stdio(false);
#ifdef Online_Judge
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif // Online_Judge
    int n = 1;
    int mi  = 0;
    while(~scanf("%d%d" , &a[n].w , &a[n].s))
    {
        a[n].num = n;
        n++;
    }
    n--;
    clr(dp , 0) , clr(path , 0);
    sort(a + 1, a + n + 1, cmp);
//    FOR(i , 0 , n)cout << a[i].w << ' ' << a[i].s << endl;
    int MAX = 0;
    dp[1] = 1;
    FORR(i , 2 , n)
    {
        int tmp = 0;
        FOR(j , 0 , i)
        {
            if(a[i].s < a[j].s && a[i].w  > a[j].w&&tmp < dp[j])
            {
                tmp = dp[j] ;
                path[i] = j;
            }
            dp[i] = tmp + 1;
        }
        if(MAX < dp[i])
        {
            MAX = dp[i];
            mi = i;
        }
    }
    vector <int> ans;
    ans.clear();
    while(path[mi])
    {
        ans.push_back(a[mi].num);
        mi = path[mi];
    }
    ans.push_back(a[mi].num);
    reverse(ans.begin() , ans.end());
    printf("%d\n" , ans.size());
    FOR(i , 0 , ans.size())
    {
        printf("%d\n" , ans[i]);
    }
    return 0;
}