Hdu3487 Play with Chain

Hdu3487 Play with Chain

 

Play with Chain Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 4980 Accepted Submission (s): 2040


Problem Description YaoYao is fond of playing his chains. He has a chain containing n diamonds on it. Diamonds are numbered from 1 to n.
At first, the diamonds on the chain is a sequence: 1, 2, 3 ,..., N.
He will perform two types of operations:
CUT a B c: He will first cut down the chain from the ath diamond to the bth diamond. And then insert it after the cth diamond on the remaining chain.
For example, if n = 8, the chain is: 1 2 3 4 5 6 7 8; We perform "CUT 3 5 4", Then we first cut down 3 4 5, and the remaining chain wocould be: 1 2 6 7 8. then we insert "3 4 5" into the chain before 5th diamond, the chain turns out to be: 1 2 6 7 3 4 5 8.

FLIP a B: We first cut down the chain from the ath diamond to the bth diamond. Then reverse the chain and put them back to the original position.
For example, if we perform "FLIP 2 6" on the chain: 1 2 6 7 3 4 5 8. The chain will turn out to be: 1 4 3 7 6 2 5 8

He wants to know what the chain looks like after perform m operations. cocould you help him?

Input There will be multiple test cases in a test data.
For each test case, the first line contains two numbers: n and m (1 ≤ n, m ≤ 3*100000 ), indicating the total number of diamonds on the chain and the number of operations respectively.
Then m lines follow, each line contains one operation. The command is like this:
CUT a B c // Means a CUT operation, 1 ≤ a ≤ B ≤ n, 0 ≤ c ≤ n-(B-a + 1 ).
FLIP a B // Means a FLIP operation, 1 ≤ a <B ≤ n.
The input ends up with two negative numbers, which shoshould not be processed as a case.

Output For each test case, you shoshould print a line with n numbers. The ith number is the number of the ith diamond on the chain.
Sample Input

8 2CUT 3 5 4FLIP 2 6-1 -1

Sample Output

1 4 3 7 6 2 5 8

Source 2010 ACM-ICPC Multi-University Training Contest (5) -- Host by BJTU
Recommend zhengfeng | We have carefully selected several similar problems for you: 3486 3479 3480 3481

 

 

 

 

Obviously, it is the Splay maintenance interval. However, this question does not automatically filter the space at the end of the line. It seems that it is really a pitfall... (on OIer's mentality of making ACM questions 2333)

 

 

 

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       #define MAXN 300005#define key c[c[rt][1]][0]using namespace std;int flag,n,m,rt,x,y,z,c[MAXN][2],s[MAXN],fa[MAXN];bool rev[MAXN];char ch[10];inline int read(){int ret=0,flag=1;char ch=getchar();while (ch<'0'||ch>'9'){if (ch=='-') flag=-1;ch=getchar();}while (ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}return ret*flag;}inline void build(int x,int y,int f){if (x>y) return;int mid=(x+y)>>1;s[mid]=1;fa[mid]=f;s[mid]=y-x+1;c[f][mid>f]=mid;if (x==y) return;build(x,mid-1,mid);build(mid+1,y,mid);}inline void pushup(int k){s[k]=s[c[k][0]]+s[c[k][1]]+1;}inline void update(int k){if (!k) return;rev[k]^=1;swap(c[k][0],c[k][1]);}inline void pushdown(int k){if (!k||!rev[k]) return;update(c[k][0]);update(c[k][1]);rev[k]=0;}inline int find(int k,int x){pushdown(k);if (s[c[k][0]]+1==x) return k;else if (s[c[k][0]]>=x) return find(c[k][0],x);else return find(c[k][1],x-s[c[k][0]]-1);}inline void rotate(int x,int &k){int y=fa[x],z=fa[y],l=(c[y][1]==x),r=l^1;if (y==k) k=x;else if (c[z][0]==y) c[z][0]=x;else c[z][1]=x;fa[x]=z;fa[y]=x;fa[c[x][r]]=y;c[y][l]=c[x][r];c[x][r]=y;pushup(y);}inline void relax(int x,int k){if (x!=k) relax(fa[x],k);pushdown(x);}inline void splay(int x,int &k){relax(x,k);while (x!=k){int y=fa[x],z=fa[y];if (y!=k){if ((c[y][0]==x)^(c[z][0]==y)) rotate(x,k);else rotate(y,k);}rotate(x,k);}pushup(x);}inline void split(int l,int r){int x=find(rt,l-1),y=find(rt,r+1);splay(x,rt);splay(y,c[rt][1]);}inline void solvecut(int l,int r,int x){split(l,r);int tmp=key;key=0;pushup(c[rt][1]);pushup(rt);split(x+1,x);key=tmp;fa[tmp]=c[rt][1];pushup(c[rt][1]);pushup(rt);}inline void solverev(int l,int r){split(l,r);update(key);pushup(c[rt][1]);pushup(rt);}inline void getans(int k){if (!k) return;pushdown(k);getans(c[k][0]);if (k!=1&&k!=n+2){if (flag++) printf(" %d",k-1);else printf("%d",k-1);}getans(c[k][1]);}int main(){n=read();m=read();while (n>=0&&m>=0){rt=0;memset(c,0,sizeof(c));memset(s,0,sizeof(s));memset(fa,0,sizeof(fa));memset(rev,0,sizeof(rev));build(1,n+2,0);rt=(n+2)>>1;F(i,1,m){scanf("%s",ch);x=read()+1;y=read()+1;if (ch[0]=='C'){z=read()+1;solvecut(x,y,z);}else solverev(x,y);}flag=0;getans(rt);printf("\n");n=read();m=read();}}