zoukankan      html  css  js  c++  java
  • Colorful Lecture Note

    Colorful Lecture Note

    时间限制:10000ms
    单点时限:1000ms
    内存限制:256MB

    描述

    Little Hi is writing an algorithm lecture note for Little Ho. To make the note more comprehensible, Little Hi tries to color some of the text. Unfortunately Little Hi is using a plain(black and white) text editor. So he decides to tag the text which should be colored for now and color them later when he has a more powerful software such as Microsoft Word.

    There are only lowercase letters and spaces in the lecture text. To mark the color of a piece of text, Little Hi add a pair of tags surrounding the text, <COLOR> at the beginning and </COLOR> at the end where COLOR is one of "red", "yellow" or "blue".

    Two tag pairs may be overlapped only if one pair is completely inside the other pair. Text is colored by the nearest surrounding tag. For example, Little Hi would not write something like "<blue>aaa<yellow>bbb</blue>ccc</yellow>". However "<yellow>aaa<blue>bbb</blue>ccc</yellow>" is possible and "bbb" is colored blue.

    Given such a text, you need to calculate how many letters(spaces are not counted) are colored red, yellow and blue.

    输入

    Input contains one line: the text with color tags. The length is no more than 1000 characters.

    输出

    Output three numbers count_red, count_yellow, count_blue, indicating the numbers of characters colored by red, yellow and blue.

    样例输入
    <yellow>aaa<blue>bbb</blue>ccc</yellow>dddd<red>abc</red>
    样例输出
    3 6 3


    package hiho_75;
    
    import java.util.ArrayList;
    import java.util.Scanner;
    import java.util.Stack;
    
    public class Main {
    
        
        
        public static void main(String[] argv){
            
            Scanner in = new  Scanner(System.in);
            String source = in.nextLine();
            in.close();
            Stack<Integer> color = new Stack<Integer>();
            char[] s = source.toCharArray();        
            int l = s.length;  int[] out = new int[3];        
            ArrayList<Integer> temp = new ArrayList<Integer>();
            ArrayList<Integer> sp_temp = new ArrayList<Integer>();
            int space_count=0;
            
            //....预处理
            for(int i=0; i<l; i++){
                int t = -i;
                char k = s[i];
                if(k=='<'){
                    if(s[i+1]=='/')
                        temp.add(t); 
                    else
                        temp.add(i);        
                    continue;
                }
                if(k=='>'){
                    temp.add(i); sp_temp.add(space_count); space_count=0;
                    continue;
                }
                    
                if(k==' ')
                    space_count++;
            }
            
            
            //......栈处理
            int L = temp.size();
            int space_n=1;
            int[] link = new int[L-1];
            for(int i=0; i<L-1; i++){            
                int j = (int) temp.get(i);
                int k = (int) temp.get(i+1);            
                if(i%2==0){
                    
                    if(j>=0){
                        int p = k-j;
                        if(p==7)
                            link[i]=-1;
                        else{
                            if(p==5)
                                link[i]=-2;    
                            else
                                link[i]=-3;
                        }
                    }
                    else{
                        j=-j;
                        int p = k-j;
                        if(p==8)
                            link[i]=-4;
                        else{
                            if(p==6)
                                link[i]=-5;    
                            else
                                link[i]=-6;
                        }
                        
                        
                    }
                }
                else{
                    if(k>0)
                        link[i]=k-j-1-sp_temp.get(space_n++);
                    else
                        link[i]=-k-j-1-sp_temp.get(space_n++);
                    
                }        
            }
            
            //...选择颜色
            int [] count = new int[3];
            for(int i=L-2; i>=0; i--){
                
                if(i%2==0){
                    
                    switch(link[i]){
                    
                    case -4:
                        color.push(0); break;
                    case -5: 
                        color.push(1); break;
                    case -6: 
                        color.push(2); break;
                    case -1: 
                        color.pop(); break;
                    case -2: 
                        color.pop(); break;
                    case -3: 
                        color.pop(); break;
                    
                    }
                }
                else{
                    if(!color.isEmpty()){
                        int k =(int) color.pop();
                        color.push(k);
                        switch(k){
                        case 0: out[0]=out[0]+link[i]; break;
                        case 1: out[1]=out[1]+link[i]; break;
                        case 2: out[2]=out[2]+link[i]; break;                    
                        }
                    }
                }
            }
            
            //...输出结果
            System.out.println(out[2]+" "+out[0]+" "+out[1]);
            
            
            
        }
    }
  • 相关阅读:
    利用数据库复制技术 实现MSSQL数据同步更新
    育子两篇你会教育自已的小孩吗
    hdu 1046 Gridland (找规律题)
    hdu 1022 Train Problem I (栈的操作,还水了半天)
    hdu 4022 Bombing (强大的map一对多的映射)
    POJ 1702 Eva's Balance (数论,平衡三进制)
    hdu 3951 Coin Game (博弈)
    hdu 1058 Humble Numbers (DP初步)
    hdu 2084 数塔 (DP初步)
    hdu 1056 HangOver (打表水题)
  • 原文地址:https://www.cnblogs.com/udld/p/5022706.html
Copyright © 2011-2022 走看看