HDOJ 4417 – Super Mario 線段樹or樹狀數組離線處理..

      題意:

                同上

      題解:

                抓著這題作死的搞~~是因為今天練習賽的一道題.SPOJ
KQUERY.直到我用最後一種樹狀數組通過了HDOJ這題後..交SPOJ的才沒逾時..看排名...時間能排到11名了..有些叼...看下時間效率..自下而上: 劃分樹、線段樹、樹狀數組、最佳化後的樹狀數組...

                劃分樹的效率最低...看來劃分樹的應用範圍還是是很有局限性...只在求kth number的時候給力..逆過來求就已經力不從心了...

                線段樹及樹狀數組處理本題..需要把詢問都存下來...按照詢問數從小到大按排個序..並且把每個數以及其序號存下來..按數字從小到大排個序...首先這一列數都是空的(全0)..然後邊放數邊統計結果...

Program:  線段樹171MS

#include<iostream>#include<stdio.h>#include<string.h>#include<cmath>#include<queue>#include<stack>#include<set>#include<map>#include<algorithm>#define ll long long#define eps 1e-5#define oo 1000000007#define pi acos(-1.0)#define MAXN 100005using namespace std;struct node{       int x,w;  }p[MAXN];struct question{       int l,r,x,id; }q[200005];int sum[MAXN<<2],ans[200005];void update(int p,int x,int l,int r,int now){       if (l==r) { sum[now]=x; return; }       int mid=l+r>>1;       if (p<=mid) update(p,x,l,mid,now<<1);       if (p>mid)  update(p,x,mid+1,r,now<<1|1);       sum[now]=sum[now<<1]+sum[now<<1|1];       return; }int query(int L,int R,int l,int r,int now){       if (L<=l && R>=r) return sum[now];       int mid=l+r>>1,ans=0;       if (L<=mid) ans+=query(L,R,l,mid,now<<1);       if (R>mid)  ans+=query(L,R,mid+1,r,now<<1|1);              return ans;}bool cmp1(node a,node b) { return a.x<b.x; }bool cmp2(question a,question b) { return a.x<b.x; } int main(){                   int n,m,i,x,t,h,T,cases;         scanf("%d",&T);       for (cases=1;cases<=T;cases++)       {               printf("Case %d:\n",cases);               scanf("%d%d",&n,&m);               for (i=1;i<=n;i++) scanf("%d",&p[i].x),p[i].w=i;               sort(p+1,p+1+n,cmp1);               p[n+1].x=oo;                for (i=1;i<=m;i++)                {                     scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].x);                     q[i].l++,q[i].r++;                     q[i].id=i;               }               sort(q+1,q+1+m,cmp2);               memset(sum,0,sizeof(sum));               h=x=1;               while (x<=n && h<=m)               {                     t=p[x].x;                     while (p[x].x==t) update(p[x].w,1,1,n,1),x++;                     while (h<=m && q[h].x<t) ans[q[h].id]=0,h++;                     while (h<=m && q[h].x<p[x].x)                      ans[q[h].id]=query(q[h].l,q[h].r,1,n,1),h++;               }               for (i=1;i<=m;i++) printf("%d\n",ans[i]);       }       return 0;}

Program: 樹狀數組 125MS

#include<iostream>#include<stdio.h>#include<string.h>#include<cmath>#include<queue>#include<stack>#include<set>#include<map>#include<algorithm>#define ll long long#define eps 1e-5#define oo 1000000007#define pi acos(-1.0)#define MAXN 100005using namespace std;struct node{       int x,w;  }p[MAXN];struct question{       int l,r,x,id; }q[200005];int sum[MAXN],ans[200005],n; void insert(int x,int k){       while (k<=n)       {             sum[k]+=x;             k+=k&(-k);       }       return;}int query(int k){       int ans=0;       while (k)       {               ans+=sum[k];               k-=k&(-k);       }       return ans;}bool cmp1(node a,node b) { return a.x<b.x; }bool cmp2(question a,question b) { return a.x<b.x; } int main(){                   int m,i,x,t,h,T,cases;        scanf("%d",&T);       for (cases=1;cases<=T;cases++)       {               printf("Case %d:\n",cases);               scanf("%d%d",&n,&m);               for (i=1;i<=n;i++) scanf("%d",&p[i].x),p[i].w=i;               sort(p+1,p+1+n,cmp1);               p[n+1].x=oo;                for (i=1;i<=m;i++)                {                     scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].x);                     q[i].l++,q[i].r++;                     q[i].id=i;               }               sort(q+1,q+1+m,cmp2);               memset(sum,0,sizeof(sum));               h=x=1;               while (x<=n && h<=m)               {                     t=p[x].x;                     while (p[x].x==t) insert(1,p[x].w),x++;                     while (h<=m && q[h].x<t) ans[q[h].id]=0,h++;                     while (h<=m && q[h].x<p[x].x)                          ans[q[h].id]=query(q[h].r)-query(q[h].l-1),h++;               }               for (i=1;i<=m;i++) printf("%d\n",ans[i]);       }       return 0;}

Program: 樹狀數組 78MS

#include<iostream>#include<stdio.h>#include<string.h>#include<cmath>#include<queue>#include<stack>#include<set>#include<map>#include<algorithm>#define ll long long#define eps 1e-5#define oo 1000000007#define pi acos(-1.0)#define MAXN 100005#define MAXM 200005using namespace std;struct node{       int x,w;  }p[MAXN]; int sum[MAXN],ans[200005],n,ql[MAXM],qr[MAXM],qx[MAXM],qid[MAXM]; void insert(int x,int k){       while (k<=n)       {             sum[k]+=x;             k+=k&(-k);       }       return;}int query(int k){       int ans=0;       while (k)       {               ans+=sum[k];               k-=k&(-k);       }       return ans;}int input(){       char c;       do       {             c=getchar();       }while (c<'0' || c>'9');       int d=0;       while (c>='0' && c<='9')       {               d=d*10+c-'0';               c=getchar();       }       return d;}bool cmp1(node a,node b) { return a.x<b.x; }bool cmp2(int a,int b) { return qx[a]<qx[b]; } int main(){                   int m,i,x,t,h,T,cases;         scanf("%d",&T);       for (cases=1;cases<=T;cases++)       {              printf("Case %d:\n",cases);              scanf("%d%d",&n,&m);              for (i=1;i<=n;i++) p[i].x=input(),p[i].w=i;              sort(p+1,p+1+n,cmp1);              p[n+1].x=oo;                for (i=1;i<=m;i++)               {                   ql[i]=input()+1,qr[i]=input()+1,qx[i]=input();                   qid[i]=i;               }              sort(qid+1,qid+1+m,cmp2);              memset(sum,0,sizeof(sum));              h=x=1;                while (x<=n && h<=m)              {                   t=p[x].x;                   while (p[x].x==t) insert(1,p[x].w),x++;                   while (h<=m && qx[qid[h]]<t) ans[qid[h]]=0,h++;                   while (h<=m && qx[qid[h]]<p[x].x)                    ans[qid[h]]=query(qr[qid[h]])-query(ql[qid[h]]-1),h++;              }              for (i=1;i<=m;i++) printf("%d\n",ans[i]);       }       return 0;}