UVA 12657 Boxes in a Line(双向链表)

UVA 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

题意:给你一个初始数组a[i]=i,每次有四种操作,第一种操作把x移到y前面,第二种操作把x移到y后面,第三种操作把x和y交换位置,第四种操作翻转整个序列。

思路:此题的关键就是第四种操作怎么进行,其实翻转之后操作一变为把x移到y后面,操作二变为把x移到y前面,操作三不变。

此题仍有一点需要注意的是要判断交换的时候还需要判断x和y是否是相邻的。

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include<bitset>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
const int maxn=100100;
int l[maxn],r[maxn];

void link(int a,int b){
    r[a]=b,l[b]=a;
}

int main(){
    int n,m,cas=0;
    while(scanf("%d%d",&n,&m)!=EOF){
        int flag=0;
        int x,y,op;
        for(int i=1;i<=n;i++)
            r[i]=i+1,l[i]=i-1;
        r[0]=1,l[0]=-1;r[n+1]=-1,l[n+1]=n;
        for(int i=1;i<=m;i++){
            scanf("%d",&op);
            if(op==4){
                flag^=1;
                continue;
            }
            if(flag&&op<=2)
                op=3-op;
            scanf("%d%d",&x,&y);
            if(op==1&&l[y]==x)  //这两步不能缺少
                continue;
            if(op==2&&r[y]==x)
                continue;
            int lx=l[x],rx=r[x],ly=l[y],ry=r[y];
            if(op==1)  //将x插入到y的前面
                link(lx,rx),link(ly,x),link(x,y);
            else if(op==2) //将x插入到y的后面
                link(lx,rx),link(y,x),link(x,ry);
            else{
                if(rx==y)
                    link(lx,rx),link(y,x),link(x,ry);
                else if(l[x]==y)
                    link(lx,rx),link(ly,x),link(x,y);
                else
                    link(lx,y),link(y,rx),link(ly,x),link(x,ry);
            }
        }
        int k=0;
        long long ans=0;
        for(int i=1;i<=n;i++){
            k=r[k];
            if(i&1)
                ans+=k;
        }
        if(flag)
            ans=(long long)n*(n+1)/2-ans;
        printf("Case %d: %lld\n",++cas,ans);
    }
    return 0;
}