UVA 12657 Boxes in a Line(双向链表)

最新推荐文章于 2024-11-11 22:17:51 发布

原创最新推荐文章于 2024-11-11 22:17:51 发布 · 278 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法 #ACM

编程成长同时被 2 个专栏收录

59 篇文章

订阅专栏

UVA

31 篇文章

订阅专栏

本文介绍了一道关于模拟指令操作的编程题，通过双向链表实现对盒子位置的操作，包括移动和翻转等，并最终计算奇数位置上的盒子总和。文章详细解释了解题思路及代码实现。

Question：
12657 Boxes in a Line
You have n boxes in a line on the table numbered 1 … n from left to right. Your task is to simulate 4
kinds of commands:
• 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y )
• 2 X Y : move box X to the right to Y (ignore this if X is already the right of Y )
• 3 X Y : swap box X and Y
• 4: reverse the whole line.
Commands are guaranteed to be valid, i.e. X will be not equal to Y .
For example, if n = 6, after executing 1 1 4, the line becomes 2 3 1 4 5 6. Then after executing
2 3 5, the line becomes 2 1 4 5 3 6. Then after executing 3 1 6, the line becomes 2 6 4 5 3 1.
Then after executing 4, then line becomes 1 3 5 4 6 2
Input
There will be at most 10 test cases. Each test case begins with a line containing 2 integers n, m
(1 ≤ n, m ≤ 100, 000). Each of the following m lines contain a command.
Output
For each test case, print the sum of numbers at odd-indexed positions. Positions are numbered 1 to n
from left to right.
Sample Input
6 4
1 1 4
2 3 5
3 1 6
4
6 3
1 1 4
2 3 5
3 1 6
100000 1
4
Sample Output
Case 1: 12
Case 2: 9
Case 3: 2500050000
题目大意：给你1~n盒子的顺序，给你指令1xy表示你要将x移到y的左边；2xy表示你要将x移到y的右边；3xy表示swap（x，y）；4表示翻转整条链，输出指令操作后的奇数位的和
解题思路：此题用双向链表，保存每个数的左右分别是那个数字。操作4执行奇数次相当于执行了1次（此时若执行操作1则相当于执行操作2，执行操作2相当于执行操作1，把op=3-op就行了），偶数次相当于并没有执行操作4
(https://www.bnuoj.com/v3/external/126/12657.pdf)

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

using namespace std;
typedef long long LL;
const int maxn=1e5+5;
int right_[maxn],left_[maxn],n,m,ncase;
void link(int l,int r)
{
    right_[l]=r;
    left_[r]=l;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
        {
            left_[i]=i-1;
            right_[i]=(i+1)%(n+1);
        }
        right_[0]=1;left_[0]=n;
        int op,x,y,inv=0;
        for(int i=0;i<m;i++)
        {
            scanf("%d",&op);
            if(op==4)
                inv=!inv;
            else
            {
                scanf("%d%d",&x,&y);
                if(op==3&&right_[y]==x)     //对一些特殊情况进行操作
                    swap(x,y);
                if(op!=3&&inv)        //遇到翻转奇数次则将op=3-op
                    op=3-op;
                if(op==1&&x==left_[y])  continue;
                if(op==2&&x==right_[y])   continue;
                int lx=left_[x],rx=right_[x],ly=left_[y],ry=right_[y];
                if(op==1)
                {
                    link(lx,rx);
                    link(ly,x);
                    link(x,y);
                }
                else if(op==2)
                {
                    link(lx,rx);
                    link(y,x);
                    link(x,ry);
                }
                else if(op==3)
                {
                    if(right_[x]==y)
                    {
                        link(lx,y);
                        link(y,x);
                        link(x,ry);
                    }
                    else
                    {
                        link(lx,y);
                        link(y,rx);
                        link(ly,x);
                        link(x,ry);
                    }
                }
            }
        }
        int b=0;
        LL ans=0;
        for(int i=1;i<=n;i++)
        {
            b=right_[b];
            if(i%2==1)
                ans+=b;
        }
        if(inv&&n%2==0)  //只有偶数位才有影响，奇数位并无影响
            ans=(LL)n*(n+1)/2-ans;  //总数减去
        printf("Case %d: %lld\n",++ncase,ans);
    }
    return 0;
}