Problem UVA12657-Boxes in a Line（数组模拟双链表）

Problem UVA12657-Boxes in a Line

Accept: 725  Submit: 9255

Time Limit: 1000 mSec

Problem Description

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

题解：数组模拟双链表。虽然是模拟，但是操作起来并不是很简单，应用双链表比较自然，困难的是中间的各种修改，我的第一份代码用的是学C语言的时候类似指针的操作，什么pre的next是x的next，next的pre是什么什么之类的，这种操作比较容易错的就是顺序问题，一旦顺序搞错，信息就会丢失，然后就不知道连到哪了，紫书上的方法十分优秀，提前先记录下来前驱、后继，这样不会丢失信息，顺序什么的就不用考虑了，然后把连边的操作封装到函数里，这样更是简化了思考，或者说是缩短了思维长度，经过这两个操作，原来十分费劲的修改关系就变得几乎是无脑操作了，这么好的方法一定要学会。

代码完全是按照紫书写的......

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;
typedef long long LL;

const int maxn = 100000+10;
int Next[maxn],Pre[maxn];
int cnt = 1;

void Link(int p,int n){
    Next[p] = n,Pre[n] = p;
}

int main()
{
    //freopen("input.txt","r",stdin);
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        for(int i = 1;i <= n;i++){
            Next[i] = (i+1)%(n+1);
            Pre[i] = i-1;
        }
        Next[0] = 1,Pre[0] = n;
        int x,y,ope,inv = 0;
        for(int i = 1;i <= m;i++){
            scanf("%d",&ope);
            if(ope == 4) inv = !inv;
            else{
                scanf("%d%d",&x,&y);
                if(inv && (ope==1 || ope==2)) ope = 3-ope;
                if(ope==1 && Pre[y]==x) continue;
                if(ope==2 && Pre[x]==y) continue;
                if(ope==3 && Next[y]==x) swap(x,y);
                int nx = Next[x],px = Pre[x];
                int ny = Next[y],py = Pre[y];
                if(ope == 1){
                    Link(px,nx);Link(py,x);Link(x,y);
                }
                if(ope == 2){
                    Link(px,nx);Link(y,x);Link(x,ny);
                }
                if(ope == 3){
                    if(Next[x]==y){
                        Link(px,y);Link(y,x);Link(x,ny);
                    }
                    else{
                        Link(px,y);Link(y,nx);
                        Link(py,x);Link(x,ny);
                    }
                }
            }
        }
        LL ans = 0;
        int t = Next[0];
        for(int i = 1;i <= n;i++){
            if(i%2 == 1) ans += t;
            t = Next[t];
        }
        if(inv && n%2==0) ans = 1LL*n*(n+1)/2-ans;
        printf("Case %d: %lld
",cnt++,ans);
    }
    return 0;
}