uva-12657

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

题解：

给你一个从1~n的数，然后给你操作方案

• 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.

用双向链表，用着刘汝佳的模版，却无限wa

我的wa代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define P_ printf(" ");
typedef long long LL;
const int MAXN=1e5+100;
int lef[MAXN],righ[MAXN];
void link(int l,int r){
	lef[r]=l;
	righ[l]=r;
}
int main(){
	int n,m;
	int s,x,y,kase=0;
	while(~scanf("%d%d",&n,&m)){
		for(int i=1;i<=n;i++){
			lef[i]=i-1;
			righ[i]=(i+1)%(n+1);
		}
		lef[0]=n;righ[0]=1;
		int flot=0;
		while(m--){
			scanf("%d",&s);
			if(s==4){
				flot=!flot;
				continue;	
				}
				scanf("%d%d",&x,&y);
				int lx=lef[x],rx=righ[x],ly=lef[y],ry=righ[y];
				if(s!=3&&flot)s=3-s;
				/****************************************/
				if(s == 3 && righ[y] == x) swap(x, y);  
        		if(s == 1 && x == lef[y]) continue;  
        		if(s == 2 && x == righ[y]) continue;
        		/****************************************/
				if(s==1){
					link(lx,rx);link(ly,x);link(x,y);
				}
				else if(s==2){
					link(lx,rx);link(y,x);link(x,ry);
				}
				else if(s==3){
					 if(righ[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 cur=0;
			LL ans=0;
			for(int i=1;i<=n;i++){
				cur=righ[cur]; 
				if(i&1)ans+=cur;                                                                                                                                                                                                         
			}
			if(flot&&n%2==0)ans=(LL)n*(n+1)/2-ans;
			printf("Case %d: %lld
",++kase,ans);
	}
	return 0;
}

　　刘汝佳的ac代码：

#include<cstdio>  
#include<algorithm>  
using namespace std;  
  
const int maxn = 100000 + 5;  
int n, left[maxn], right[maxn];  
  
inline void link(int L, int R) {  
  right[L] = R; left[R] = L;  
}  
  
int main() {  
    freopen("1.txt","r",stdin);  
    freopen("2.txt","w",stdout);  
  int m, kase = 0;  
  while(scanf("%d%d", &n, &m) == 2) {  
    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;  
  
    while(m--) {  
      scanf("%d", &op);  
      if(op == 4) inv = !inv;  
      else {  
        scanf("%d%d", &X, &Y);  
        if(op == 3 && right[Y] == X) swap(X, Y);  
        if(op != 3 && inv) 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;  
    long long ans = 0;  
    for(int i = 1; i <= n; i++) {  
      b = right[b];  
      if(i % 2 == 1) ans += b;  
    }  
    if(inv && n % 2 == 0) ans = (long long)n*(n+1)/2 - ans;  
    printf("Case %d: %lld
", ++kase, ans);  
  }  
  return 0;  
}  

　再次比赛，出现了这个题，要注意，当两个数挨着的时候，要特判，否则会造成链表出乱，死循环，ac:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 100100;
typedef long long LL;
int L[MAXN], R[MAXN];
void link(int l, int r){
    R[l] = r;
    L[r] = l;
}
int main(){
    int n, m;
    int kase = 0;
    while(~scanf("%d%d", &n, &m)){
        for(int i = 0; i < n; i++){
            link(i, i + 1);
        }
        L[0] = n; R[n] = 0;
        int q, x, y;
        int l, r, ll, rr;
        int sw = 0;
        while(m--){
            scanf("%d", &q);
            if(q == 4){
                sw ^= 1;
            }
            else{
                scanf("%d%d", &x, &y);
                if(q == 1 || q == 2){
                    if(sw)
                        q = 3 - q;
                }
                if(q == 1 && L[y] == x)
                    continue;
                if(q == 2 && R[y] == x)
                    continue;
                if(q == 3){
                    l = L[x]; r = R[x];
                    ll = L[y]; rr = R[y];
                    if(R[x] == y){
                        link(l, y);
                        link(y, x);
                        link(x, rr);
                    }
                    else if(R[y] == x){
                        link(ll, x);
                        link(x, y);
                        link(y, r);
                    }
                    else{
                        link(l, y); link(y, r);
                        link(ll, x); link(x, rr);
                    }
                }
                else if(q == 2){
                    l = L[x]; r = R[x];
                    ll = L[y]; rr = R[y];
                    link(l, r);
                    link(y, x); link(x, rr);
                }
                else if(q == 1){
                    l = L[x]; r = R[x];
                    ll = L[y]; rr = R[y];
                    link(l, r);
                    link(ll, x); link(x, y);
                }
            }
        }
        LL ans = 0;
        if(sw){
            for(int i = L[0], j = 1; i ; i = L[i], j++){
                if(j & 1)
                    ans += i;
            }
        }
        else{
            for(int i = R[0], j = 1; i ; i = R[i], j++){
                if(j & 1)
                    ans += i;
            }
        }
        printf("Case %d: %lld
", ++kase, ans);
    }
    return 0;
}