zoukankan      html  css  js  c++  java
  • (最长上升子序列 并记录过程)FatMouse's Speed -- hdu -- 1160

    http://acm.hdu.edu.cn/showproblem.php?pid=1160

    FatMouse's Speed

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 12338    Accepted Submission(s): 5405
    Special Judge


    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
    4
    4
    5
    9
    7
    #include<iostream>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<algorithm>
    
    using namespace std;
    #define N 10005
    #define oo 0x3f3f3f3f
    
    /**
    
    感觉有点用导弹防御的思想
    每次记录从 0 到 i 的最长个数
    并用 pre[i] 记录 i 点是从哪个点过来的
    
    **/
    
    /// number 是记录它原来的位置
    
    struct node
    {
        int w, s, number;
    }a[N];
    
    /// pre[] 是个很好的记录方式,可以记录它的是从哪个过来的(它的上一步)
    
    int dp[N], pre[N];
    
    /// 先按 w 从小到大排, 再按 s 从大到小排
    int cmp(node n1, node n2)
    {
        if(n1.w!=n2.w)
           return n1.w < n2.w;
        return n1.s > n2.s;
    }
    
    int main()
    {
        int i=1, j, n;
    
        memset(a, 0, sizeof(a));
        memset(dp, 0, sizeof(dp));
    
        while(scanf("%d%d", &a[i].w, &a[i].s)!=EOF)
        {
            a[i].number = i;
            i++;
        }
    
        n=i;
        sort(a+1, a+i+1, cmp);
    
    /**
        for(i=1; i<n; i++)
            printf("%d %d %d
    ", a[i].number, a[i].w, a[i].s);
    **/
    
        for(i=1; i<n; i++)
        {
            pre[i] = i;
            dp[i] = 1;
            for(j=1; j<i; j++)
            {
                if(a[i].w!=a[j].w && a[i].s<a[j].s)
                {
                    if(dp[i]<dp[j]+1)
                    {
                         pre[i] = j;
                         dp[i] = dp[j] + 1;
                    }
                }
            }
        }
    
        int Max=-oo, index=0;
    
        for(i=1; i<n; i++)
        {
            if(dp[i]>Max)
            {
                Max = dp[i];
                index = i;
            }
        }
    
    
        printf("%d
    ", Max);
        i=0;
        int b[N]={0};
        while(pre[index]!=index)
        {
            b[i++] = a[index].number;
            index = pre[index];
        }
    
        printf("%d
    ", a[index].number);
        for(j=i-1; j>=0; j--)
            printf("%d
    ", b[j]);
    
        return 0;
    }
    View Code
    勿忘初心
  • 相关阅读:
    arcgis要素转json
    GeoJSON.js:221 Uncaught Error: Invalid GeoJSON object.
    深入学习SAP UI5框架代码系列之一:UI5 Module的懒加载机制
    SAP Spartacus简介
    金庸逝世两周年纪念日:一个失意程序员的呓语
    一个用于SAP UI5学习的脚手架应用,没有任何后台API的依赖
    通过最简单的button控件,深入学习SAP UI5框架代码系列之零
    如何使用SAP UI5 SDK网站查询指定控件的属性如何使用
    一行代码将SAP CDS view数据以ALV的方式输出
    演讲预告:一个月的住院经历,我悟到了哪些和程序员职场发展相关的心得
  • 原文地址:https://www.cnblogs.com/YY56/p/4864945.html
Copyright © 2011-2022 走看看