UVA12657解题报告
2017-06-30 22:33
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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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