可以直接参考hiho一下的详细解答：http://hihocoder.com/discuss/question/2736/
下面是我的解题报告：
1.该题主要是统计三种颜色的字母的个数，red，yellow，blue；
2.需要注意的是，必须要判断字符是否为字母的情况，题目提到空格不加入统计。
3.思路：
使用栈存储和消除颜色，如果遇到<XXX>，则压入栈，如果遇到</XXX>，则pop出栈顶的元素，字母的颜色即栈顶的颜色。如果栈为空，则字母不统计。
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
AC代码：
[c language=”++”]
//#include<string>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include <iomanip>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;
int getTagColor(string &str, int start)
{//红色返回0，黄色返回1，蓝色返回0
if (start < str.size())
{
if (start < str.size())
{
if (str[start] == ‘r’&&
start + 2 < str.size() &&
str[start + 1] == ‘e’&&
str[start + 2] == ‘d’)
return 0;
else if (str[start] == ‘y’&&
start + 5 < str.size() &&
str[start + 1] == ‘e’&&
str[start + 2] == ‘l’&&
str[start + 3] == ‘l’&&
str[start + 4] == ‘o’&&
str[start + 5] == ‘w’)
return 1;
else if (str[start] == ‘b’&&
start + 3 < str.size() &&
str[start + 1] == ‘l’&&
str[start + 2] == ‘u’&&
str[start + 3] == ‘e’)
return 2;
else return -4;
}
else return -4;
}
else return -4;
}
int getTag(string &str, int start)
{//获得标签的属性即颜色，属性包括颜色是开始还是结束
if (start < str.size())
{
if (str[start] == ‘<‘)
{
//颜色结束
if (start + 1 < str.size() && str[start + 1] == ‘/’)
return getTagColor(str, start + 2) + 3;
//颜色开始
else
return getTagColor(str, start + 1);
}
else return -1;
}
else return -1;
}
int main(void)
{
string text;
getline(cin , text);
stack<int> colorTag;
int times[3] = { 0 };
for (int i = 0; i < text.size(); i++)
{
int tag = getTag(text, i);
if (tag >= 0)
{
if (tag >= 3 && !colorTag.empty())
{//为颜色结束
colorTag.pop();
if (tag == 3) i += (3 + 2); //红色，跳过5个字符，包括’/’和最后的’>’
else if (tag == 4) i += (6 + 2);//黄色，跳过8个字符
else if (tag == 5) i += (4 + 2);//蓝色，跳过6个字符
}
else if (tag >= 0)
{
colorTag.push(tag);
if (tag == 0) i += (3+1); //红色，跳过3个字符，包括最后的’>’
else if (tag == 1) i += (6+1);//黄色，跳过6个字符
else if (tag == 2) i += (4+1);//蓝色，跳过4个字符
}
}
else if (!colorTag.empty() &&( (text[i] >= ‘a’&&text[i] <= ‘z’) || (text[i] >= ‘A’&&text[i] <= ‘Z’)))
{//栈不为空，需要注意判断为字母，题目提到空格不统计
times[colorTag.top()]++;
}
}
printf("%d %d %d\n", times[0],times[1], times[2]);
return 0;
}
[/c]