UVA 12657 链表模拟

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1~n个箱子，每个箱子的价值为1~n,进行m次操作,共有四种类型的操作,

1 x y表示把x放在y的左端；

2 x y 表示把x放在y的右端;；

3 x y表示把x和y互换位置；

4 表示把全部箱子的位置倒转过来

问你在m次操作之后，奇数位置的箱子价值总和为多少。

从插入等操作，可以看出这道题是链表问题，用STL的链表尝试，发现超时了，最终用数组模拟链表A了这道题。

可以看出STL的链表比自己写的链表速度要慢。

以下是超时代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<list>
#include<algorithm>
using namespace std;

list<int>l;
list<int>::iterator it1,it2,it;

void solve(int a,int a1=1,int a2=1){
if(a==4){l.reverse();return ;}
if(a1==a2)return ;
it1=find(l.begin(),l.end(),a1);
it2=find(l.begin(),l.end(),a2);
if(a==1){
l.erase(it1);
l.insert(it2,a1);
}
if(a==2){
l.erase(it1);
l.insert(++it2,a1);
}
if(a==3){
l.insert(it2,a1);
l.insert(it1,a2);
l.erase(it1);
l.erase(it2);
}
}

int main()
{
int m,n,flag,x,y,Case=1;
while(~scanf("%d%d",&n,&m)){
l.clear();
for(int i=1;i<=n;i++){
l.push_back(i);
}
for(int i=0;i<m;i++){
scanf("%d",&flag);
if(flag==4){
solve(flag);
}
else{
scanf("%d%d",&x,&y);
solve(flag,x,y);
}
}
long long ans=0;
for(it=l.begin();it!=l.end();++it){
ans+=*it;
if(++it==l.end())break;
}
printf("Case %d: %I64d\n",Case++,ans);
}

return 0;
}

数组模拟双向链表操作的代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100010;
int l[maxn],r[maxn];
void unlink(int p1){
int t1=l[p1];
int t2=r[p1];
r[t1]=t2;
l[t2]=t1;
}
void link(int p1,int p2,int v){
r[p1]=v;
r[v]=p2;
l[p2]=v;
l[v]=p1;
}
int main()
{
int n,m;
int a,b,c;
int Case = 1;
while(~scanf("%d%d",&n,&m)){
memset(r,0,sizeof(r));
memset(l,0,sizeof(l));
for(int i=1;i<=n;i++){
r[i]=(i+1)%(n+1);
l[i]=i-1;
}
r[0]=1;
l[0]=n;
bool flag=true;
for(int i=0;i<m;i++){
scanf("%d",&a);
if(a==4){flag=!flag;continue;};
scanf("%d%d",&b,&c);
if(a==1){
if(b==l[c])continue;
unlink(b);
if(flag)link(l[c],c,b);
else link(c,r[c],b);
}
if(a==2){
if(b==r[c])continue;
unlink(b);
if(flag) link(c,r[c],b);
else link(l[c],c,b);
}
if(a==3){
if(r[b]==c){
int t1=l[b];
int t2=r[c];
l[b]=c;
r[c]=b;
l[c]=t1;
r[b]=t2;
r[t1]=c;
l[t2]=b;
}
else if(l[b]==c){
int t1=l[c];
int t2=r[b];
l[c]=b;
r[b]=c;
l[b]=t1;
r[c]=t2;
r[t1]=b;
l[t2]=c;
}
else{
unlink(b);
unlink(c);
int t1=l[c];
int t2=r[c];
link(l[b],r[b],c);
link(t1,t2,b);
}
}
}
long long ans=0;
if(flag){
bool flag1=true;
for(int i=r[0];i!=0;i=r[i]){
if(flag1){
ans+=i;
}
flag1=!flag1;
}
}
else{
bool flag1=true;
for(int i=l[0];i!=0;i=l[i]){
if(flag1)ans+=i;
flag1=!flag1;
}
}
printf("Case %d: %lld\n",Case++,ans);
}

return 0;
}