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UVA12657解题报告

2017-06-30 22:33 281 查看
AC代码 Time 80ms
#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn=100000+10;
int left[maxn],right[maxn],inv;

void link(int L,int R)
{
left[R]=L;
right[L]=R;
}

long long number(int n)
{
int x=0;
long long total=0;
for(int i=1;i<=n;i++)
{
x=right[x];
if(i%2) total+=x;
}
if(inv && n%2==0) total=(long long)n*(n+1)/2-total;
return total;
}

int main()
{
int n,m,T=1;
while(scanf("%d %d",&n,&m)==2)
{
//init
for(int i=1;i<=n;i++)
{
left[i]=i-1;
right[i]=(i+1)%(n+1);
}
right[0]=1; left[0]=n;
inv=0;

while(m--)
{
int order,X,Y;
scanf("%d",&order);
if(order==4)
{
inv=!inv;
continue;
}
scanf("%d %d",&X,&Y);
if(order==3 && right[Y]==X) swap(X,Y);  //X,Y相邻时要进行特殊考虑
if(order!=3 && inv) order=3-order;
if(order==1 && left[Y]==X) continue;
if(order==2 && right[Y]==X) continue;

int LX,RX,LY,RY;
LX=left[X]; RX=right[X]; LY=left[Y]; RY=right[Y];
if(order==1)
{
link(X,Y); link(LX,RX); link(LY,X);
}
else if(order==2)
{
link(Y,X); link(LX,RX); link(X,RY);
}
else if(order==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); }
}
}
printf("Case %d: %lld\n",T++,number(n));
}
return 0;
}
开始用链表做的超限代码(实际上也是错误的)
#include<cstdio>
#include<cstdlib>
using namespace std;

int reverse;

typedef struct Node{
int id;
struct Node *pre,*next;
}Node,*LNode;

void init(LNode &head,int n,LNode &rear){
head=(LNode)malloc(sizeof(Node));
head->id=-1; head->pre=NULL;
rear=(LNode)malloc(sizeof(Node));
rear->id=-1; rear->next=NULL;
head->next=rear; rear->pre=head;
for(int i=1;i<=n;i++){
LNode p=(LNode)malloc(sizeof(Node));
p->next=rear; p->pre=rear->pre;
rear->pre->next=p; rear->pre=p;
p->id=i;
}
}

int locate(LNode L,LNode &p,LNode &q,int x,int y){
LNode k=L->next;
int num=0;
while(k->id!=-1 && num!=2){
if(k->id==x) { p=k; num++; }
if(k->id==y) { q=k; num++; }
k=k->next;
}
}

void right(LNode &head,int x,int y){
LNode p,q;
locate(head,p,q,x,y);
if(q->next==p) return;  //if x is left to y,return!
p->next->pre=p->pre; p->pre->next=p->next;
q->next->pre=p; p->next=q->next; p->pre=q; q->next=p; //insert p into the q left
}
void left(LNode &head,int x,int y){
LNode p,q;
locate(head,p,q,x,y);
if(q->pre==p) return; //if x is right to y,return!
p->next->pre=p->pre; p->pre->next=p->next; //split p from chain
q->pre->next=p; p->next=q; p->pre=q->pre; q->pre=p;
}

void replace(LNode &head,int x,int y){
LNode p,q,k;
locate(head,p,q,x,y);
int temp=p->id;
p->id=q->id;
q->id=temp;
}
long long sum(LNode head,LNode rear){
int i=1;
long long ans=0;
LNode p;
if(!reverse){
p=head->next;
while(p->id!=-1){
if(i%2) ans+=p->id;
i++; p=p->next;
}
}
else{
p=rear->pre;
while(p->id!=-1){
if(i%2) ans+=p->id;
i++; p=p->pre;
}
}
return ans;
}

void print(LNode head){
LNode p=head->next;
while(p->id!=-1){
printf("%d  ",p->id);
p=p->next;
}
printf("\n");
}

int main(){
int N,M,Test=1;
while(scanf("%d %d",&N,&M)==2){
LNode head,rear;
init(head,N,rear);
reverse=0;
for(int i=1;i<=M;i++){
int order;
scanf("%d",&order);
if(order==4) { reverse=!reverse; continue; }
int x,y;
scanf("%d %d",&x,&y);
switch(order){
case 1:
if(reverse) right(head,x,y);
else left(head,x,y);
break;
case 2:
if(reverse) left(head,x,y);
else right(head,x,y);
break;
case 3:
replace(head,x,y);
break;
}
}
printf("Case %d: %lld\n",Test++,sum(head,rear));
}
return 0;
}
错误的代码,1,2操作没有考虑到已经排好序的情况,3没有考虑X,Y相邻的情况,代码下面给出一组对于3操作出错的数据
#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn=100000+10;
int left[maxn],right[maxn],inv;

void link(int L,int R)
{
left[R]=L;
right[L]=R;
}

long long number(int n)
{
int x=0;
long long total=0;
for(int i=1;i<=n;i++)
{
x=right[x];
if(i%2) total+=x;
}
if(inv && n%2==0) total=(long long)n*(n+1)/2-total;
return total;
}

int main()
{
int n,m,T=1;
while(scanf("%d %d",&n,&m)==2)
{
//init
for(int i=1;i<=n;i++)
{
left[i]=i-1;
right[i]=(i+1)%(n+1);
}
right[0]=1; left[0]=n;
inv=0;

while(m--)
{
int order,X,Y;
scanf("%d",&order);
if(order==4)
{
inv=!inv;
continue;
}
scanf("%d %d",&X,&Y);
if(order!=3 && inv) order=3-order;

int LX,RX,LY,RY;
LX=left[X]; RX=right[X]; LY=left[Y]; RY=right[Y];
if(order==1)
{
link(X,Y); link(LX,RX); link(LY,X);
}
else if(order==2)
{
link(Y,X); link(LX,RX); link(X,RY);
}
else if(order==3)
{
link(LX,Y); link(Y,RX); link(LY,X); link(X,RY);
}
}
printf("Case %d: %lld\n",T++,number(n));
}
return 0;
}

/*错误数据
4 1
3 2 3
正确答案 3*/
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