Gym - 101620H_Hidden Hierarchy(树+模拟)

Hidden Hierarchy

题目链接

题目描述

You are working on the user interface for a simple text-based file explorer. One of your tasks is to build a navigation pane displaying the directory hierarchy. As usual, the filesystem consists of directories which may contain files and other directories, which may, in turn, again contain files and other directories etc. Hence, the directories form a hierarchical tree structure. The top-most directory in the hierarchy is called the root directory. If directory d directly contains directory e we will say that d is the parent directory of e while e is a subdirectory od d. Each file has a size expressed in bytes. The directory size is simply the total size of all files directly or indirectly contained inside that directory.
All files and all directories except the root directory have a name — a string that always starts with a letter and consists of only lowercase letters and “.” (dot) characters. All items (files and directories) directly inside the same parent directory must have unique names. Each item (file and directory) can be uniquely described by its path — a string built according to the following rules:

• Path of the root directory is simply “/” (forward slash).
• For a directory d, its path is obtained by concatenating the directory names top to bottom along the hierarchy from the root directory to d, preceding each name with the “/” character and placing another “/” character at the end of the path.
• For a file f , its path is the concatenation of the parent directory path and the name of file f .
  We display the directory hierarchy by printing the root directory. We print a directory d by outputting a line of the form “md pd sd” where pd and sd are the path and size of directory d respectively, while md is its expansion marker explained shortly. If d contains other directories we must choose either to collapse it or to expand it. If we choose to expand d we print (using the same rules) all of its subdirectories in lexicographical order by name. If we choose to collapse directory d, we simply ignore its contents.
  The expansion marker md is a single blank character when d does not have any subdirectories, “+” (plus) character when we choose to collapse d or a “-” (minus) character when we choose expand d.
  Given a list of files in the filesystem and a threshold integer t, display the directory hierarchy ensuring that each directory of size at least t is printed. Additionally, the total number of directories printed should be minimal. Assume there are no empty directories in the filesystem — the entire hierarchy can be deduced from the provided file paths. Note that the root directory has to be printed regardless of its size. Also note that a directory of size at least t only has to be printed, but not necessarily expanded.

输入

The first line contains an integer n (1 ≤ n ≤ 1 000) — the number of files. Each of the following n lines contains a string f and an integer s (1 ≤ s ≤ 1e6) — the path and the size of a single file. Each path is at most 100 characters long and is a valid file path according to the rules above. All paths will be different.
The following line contains an integer t (1 ≤ t ≤ 1e9) — the threshold directory size.

输出

Output the minimal display of the filesystem hierarchy for the given threshold as described above.

样例

input

9
/sys/kernel/notes 100
/cerc/problems/a/testdata/in 1000000
/cerc/problems/a/testdata/out 8
/cerc/problems/a/luka.cc 500
/cerc/problems/a/zuza.cc 5000
/cerc/problems/b/testdata/in 15
/cerc/problems/b/testdata/out 4
/cerc/problems/b/kale.cc 100
/cerc/documents/rules.pdf 4000
10000

output

- / 1009727
- /cerc/ 1009627
  /cerc/documents/ 4000
- /cerc/problems/ 1005627
- /cerc/problems/a/ 1005508
  /cerc/problems/a/testdata/ 1000008
+ /cerc/problems/b/ 119
+ /sys/ 100

input

8
/b/test/in.a 100
/b/test/in.b 1
/c/test/in.a 100
/c/test/in.b 1
/c/test/pic/in.a.svg 10
/c/test/pic/in.b.svg 10
/a/test/in.a 99
/a/test/in.b 1
101

output

- / 322
+ /a/ 100
- /b/ 101
  /b/test/ 101
- /c/ 121
+ /c/test/ 121

input

2
/a/a/a 100
/b.txt 99
200

output

+ / 199

题意

这个题目超级长，废话也超级多。大概就是做一个目录。
有n条路径，每条路径下有一个大小为val的文件。
根文件为“/”；
给一个t；
如果目录下文件大小有一个大于等于t，展开“-”；
如果都小于t，折叠“+”；
如果没有文件，则输出一个“ ”。
注意：输出路径按字典序排序。

思路

首先把每个文件路径及其大小读入，按字典序排序。
然后把路径分开，建树。
根据题目要求输出就可以了。
有几个点注意：
内存问题，刚开始用数组存储，结果导致RE跟MLE二选一，后来改成了vector。
输出问题，这里需要采用递归输出。
剩下的就是疯狂模拟就可以了。

附上AC代码。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>

using namespace std;

struct WJ//记录文件，及其目录下有多少文件；
{
    int len,w,parent;
    char name[105];
    vector<int>next;
};
struct Char//记录输入的字符串；
{
    char s[105];
    int w;
};

vector<WJ>p;
char name[105][105];
int num,w,num2;//num，文件个数；num2，每个路径的文件个数。

bool cmp(Char a,Char b)
{
    if(strcmp(a.s,b.s)>0)
        return false;
    else if(strcmp(a.s,b.s)==0)
        return a.w<b.w;
    return true;
}

void build(int x,int w,int j)
{
    p[x].w += w;
    if(j>=num2-1)//递归到路径文件就可以退出了；
        return;
    int i,a;
    for(i=0;i<p[x].len;i++)
    {
        a = p[x].next[i];
        if(strcmp(p[a].name,name[j])==0)
        {
            build(a,w,j+1);
            return;
        }
    }
    WJ q;
    q.len = q.w = 0;
    q.parent = x;
    strcpy(q.name,name[j]);
    p[x].next.push_back(num++);
    p[x].len++;
    p.push_back(q);
    build(num-1,w,j+1);
}

void print(int x)//递归输出文件路径
{
    if(p[x].parent!=-1)
        print(p[x].parent);
    if(p[x].parent==-1)
        return;
    printf("%s/",p[x].name);
}

void show(int x)
{
    int i,a;
    if(p[x].len==0)
    {
        printf("  /");
        print(x);
        printf(" %d\n",p[x].w);
        return;
    }
    for(i=0;i<p[x].len;i++)
    {
        a = p[x].next[i];
        if(p[a].w>=w)
            break;
    }
    if(i==p[x].len)
    {
        printf("+ /");
        print(x);
        printf(" %d\n",p[x].w);
    }
    else
    {
        printf("- /");
        print(x);
        printf(" %d\n",p[x].w);
        for(i=0;i<p[x].len;i++)
            show(p[x].next[i]);
    }
}

int main()
{
    int n,i,j,k,len;
    Char s[1050];
    WJ q;
    q.len = q.w = 0;
    q.parent = -1;
    p.push_back(q);
    scanf("%d",&n);
    for(i=0;i<n;i++)
        scanf("%s%d",s[i].s,&s[i].w);
    scanf("%d",&w);
    sort(s,s+n,cmp);
    num = 1;
    for(k=0;k<n;k++)//把每个路径分割；
    {
        num2 = 0;

        len = strlen(s[k].s);
        for(i=0;i<len;i++)
        {
            if(s[k].s[i]=='/')
            {
                for(j=i+1;s[k].s[j]!='/'&&j<len;j++)
                    name[num2][j-i-1] = s[k].s[j];
                name[num2][j-i-1] = '\0';
                num2++;
            }
            i = j - 1;
        }
        build(0,s[k].w,0);
    }
    show(0);
    return 0;
}