hdu 1160 FatMouse's Speed (最长上升子序列+打印路径)

Problem Description

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

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

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

Sample Output

主要是打印路径的一些小细节

#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);

}

巴特西

hdu 1160 FatMouse's Speed (最长上升子序列+打印路径)

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