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

FatMouse's Speed

Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

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 Specification

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.

Output Specification

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.

Sample Input

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

Output for Sample Input

4
4
5
9
7

【題意】

給出每隻倉鼠的重量和速度，求序列中重量嚴格遞增速度嚴格遞減的最長序列，並輸出該序列的長度及序列，若有多種答案輸出一種即可。

【解題思路】

先根據給定要求排個序，此時倉鼠的重量已經是非遞減序列，v[i]陣列是包含第i只倉鼠在內的最長序列，然後每一次更新dp值的時候說明第i只倉鼠的最長序列已經改變了，所以需要v[i]=v[j]，並且把自己也壓入，最後遍歷一下v取最長的序列長度即可。

【程式碼】

#include<bits/stdc++.h>
using namespace std;
struct Node
{
    int weight,speed,num;
}node[1005];
int dp[1005];
vector<int>v[1005],ans;
bool cmp(Node a,Node b)
{
    return (a.weight!=b.weight)?a.weight<b.weight:a.speed>b.speed;
}
int main()
{
    int cnt=1;
    while(~scanf("%d%d",&node[cnt].weight,&node[cnt].speed))
    {
        node[cnt].num=cnt;
        cnt++;
    }
    sort(node,node+cnt,cmp);
    for(int i=1;i<cnt;i++)
    {
        dp[i]=1;
        v[i].push_back(i);
    }
    for(int i=1;i<cnt;i++)
    {
        for(int j=i-1;j>=1;j--)
        {
            if(node[i].weight>node[j].weight && node[i].speed<node[j].speed && dp[i]<dp[j]+1)
            {
                dp[i]=dp[j]+1;
                v[i]=v[j];
                v[i].push_back(i);
            }
        }
    }
    int MAX=0;
    for(int i=1;i<cnt;i++)
    {
        if(MAX<v[i].size())
        {
            MAX=v[i].size();
            ans=v[i];
        }
    }
    printf("%d\n",ans.size());
    for(int i=0;i<ans.size();i++)
        printf("%d\n",node[ans[i]].num);
    return 0;
}