連結串列-雙向連結串列&&UVa12657 Boxes in a Line(移動盒子)的理解與解析

You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4kinds 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 executing2 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 nfrom 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

Solution

解決方法來自書上

#include <iostream>
#include <cstdio>

using namespace std;
const int maxn=100000+5;
int n,leftt[maxn],rightt[maxn];//left and right is ambiguous

void link(int L,int R)
{
    rightt[L]=R;leftt[R]=L;
}

int main()
{
    int m,kase=0;
    while(scanf("%d%d",&n,&m)==2){
        for(int i=1;i<=n;i++){
            leftt[i]=i-1;
            rightt[i]=(i+1)%(n+1);
        }
        rightt[0]=1;leftt[0]=n;
        int op,X,Y,inv=0;

        while(m--){
            scanf("%d",&op);
            if(op==4) inv=!inv;
            else{
                scanf("%d%d",&X,&Y);
                if(op==3&&rightt[Y]==X) swap(X,Y);//!為了轉化為同一種情況來處理，x、y只是一個代號，反正結果變得是值
                if(op!=3&&inv) op=3-op;
                if(op==1&&X==leftt[Y]) continue;
                if(op==2&&X==rightt[Y]) continue;

                int LX=leftt[X],RX=rightt[X],LY=leftt[Y],RY=rightt[Y];
                //下面分類討論，不要漏掉
                if(op==1){
                    link(LX,RX);link(LY,X);link(X,Y);
                }//X的左右連起來，X和Y左，X和Y
                else if(op==2){
                    link(LX,RX);link(Y,X);link(X,RY);
                }//X的左右連起來，Y和X，X和Y右    按照鏈的順序來寫，可能思路更清晰
                else if(op==3){
                    if(rightt[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;
        long long ans=0;
        for(int i=1;i<=n;i++){
            b=rightt[b];//向右推移
            if(i%2==1) ans+=b;
        }
        if(inv&&n%2==0) ans=(long long)n*(n+1)/2-ans;//如果n不是偶數的話，倒一下結果也一樣
        printf("Case %d:  %lld\n",++kase,ans);
    }
    return 0;
}

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