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array(2019CCPC网络预选赛 hdu6703主席树+set)主席树求大于等于k的最小值

發布時間:2023/12/15 编程问答 37 豆豆
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Problem Description
You are given an array a1,a2,…,an(?i∈[1,n],1≤ai≤n). Initially, each element of the array is unique.

Moreover, there are m instructions.

Each instruction is in one of the following two formats:

  • (1,pos),indicating to change the value of apos to apos+10,000,000;
  • (2,r,k),indicating to ask the minimum value which is not equal to any ai ( 1≤i≤r ) and **not less ** than k.
  • Please print all results of the instructions in format 2.

    Input
    The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.

    In each test case, there are two integers n(1≤n≤100,000),m(1≤m≤100,000) in the first line, denoting the size of array a and the number of instructions.

    In the second line, there are n distinct integers a1,a2,…,an (?i∈[1,n],1≤ai≤n),denoting the array.
    For the following m lines, each line is of format (1,t1) or (2,t2,t3).
    The parameters of each instruction are generated by such way :

    For instructions in format 1 , we defined pos=t1⊕LastAns . (It is promised that 1≤pos≤n)

    For instructions in format 2 , we defined r=t2⊕LastAns,k=t3⊕LastAns. (It is promised that 1≤r≤n,1≤k≤n )

    (Note that ⊕ means the bitwise XOR operator. )

    Before the first instruction of each test case, LastAns is equal to 0 .After each instruction in format 2, LastAns will be changed to the result of that instruction.

    (∑n≤510,000,∑m≤510,000 )

    Output
    For each instruction in format 2, output the answer in one line.

    Sample Input
    3
    5 9
    4 3 1 2 5
    2 1 1
    2 2 2
    2 6 7
    2 1 3
    2 6 3
    2 0 4
    1 5
    2 3 7
    2 4 3
    10 6
    1 2 4 6 3 5 9 10 7 8
    2 7 2
    1 2
    2 0 5
    2 11 10
    1 3
    2 3 2
    10 10
    9 7 5 3 4 10 6 2 1 8
    1 10
    2 8 9
    1 12
    2 15 15
    1 12
    2 1 3
    1 9
    1 12
    2 2 2
    1 9

    Sample Output
    1
    5
    2
    2
    5
    6
    1
    6
    7
    3
    11
    10
    11
    4
    8
    11

    Hint

    note:
    After the generation procedure ,the instructions of the first test case are :
    2 1 1, in format 2 and r=1 , k=1
    2 3 3, in format 2 and r=3 , k=3
    2 3 2, in format 2 and r=3 , k=2
    2 3 1, in format 2 and r=3 , k=1
    2 4 1, in format 2 and r=4 , k=1
    2 5 1, in format 2 and r=5 , k=1
    1 3 , in format 1 and pos=3
    2 5 1, in format 2 and r=5 , k=1
    2 5 2, in format 2 and r=5 , k=2

    the instructions of the second test case are :
    2 7 2, in format 2 and r=7 , k=2
    1 5 , in format 1 and pos=5
    2 7 2, in format 2 and r=7 , k=2
    2 8 9, in format 2 and r=8 , k=9
    1 8 , in format 1 and pos=8
    2 8 9, in format 2 and r=8 , k=9

    the instructions of the third test case are :
    1 10 , in format 1 and pos=10
    2 8 9 , in format 2 and r=8 , k=9
    1 7 , in format 1 and pos=7
    2 4 4 , in format 2 and r=4 , k=4
    1 8 , in format 1 and pos=8
    2 5 7 , in format 2 and r=5 , k=7
    1 1 , in format 1 and pos=1
    1 4 , in format 1 and pos=4
    2 10 10, in format 2 and r=10 , k=10
    1 2 , in format 1 and pos=2
    昨天卡了很久的一個題,你開始線段樹+set一直超時,后來改了主席樹,但是還是卡了很久。。
    題意:要找出一個數,這個數在1~r中不能出現過,并且還要不小于k。并且讀入的數據還都要異或上上一次的答案。
    一開始理解錯了,以為必須是數組里的數才行。但是第一個樣例中就有一個數不是數組里的數,除此之外,題目中有說所有的數都是獨一無二的。在執行操作1的時候,一個數加上1e7之后,就相當于作廢了。因為數組中所有的數字都是不大于1e5的,所以異或之后的值也是不大于1e5的。這樣的話,在加上1e7之后,就相當于這個值在1~r中刪除了,又可以再用了。那這樣我們把這樣的值存到一個set里面,r+1到n的值用主席樹去求。主席樹是好多權值線段樹的集合,是按著權值來的,存的是每個數的出現的個數。用主席樹去求r+1 ~ n中大于等于k最小的數。
    代碼如下:

    #include<bits/stdc++.h> #define ll long long #define inf 0x3f3f3f3f using namespace std;const int maxx=1e5+100; struct node{int l;int r;int sum; }p[maxx<<5]; int a[maxx]; int root[maxx]; int n,m,tot;inline void pushup(int cur) {p[cur].sum=p[p[cur].l].sum+p[p[cur].r].sum; } inline int build(int l,int r) {int cur=++tot;p[cur].sum=0;if(l==r) return cur;int mid=l+r>>1;build(l,mid);build(mid+1,r);return cur; } inline int update(int rot,int l,int r,int pos) {int cur=++tot;p[cur]=p[rot];if(l==r) {p[cur].sum++;return cur;}int mid=l+r>>1;if(pos<=mid) p[cur].l=update(p[rot].l,l,mid,pos);else p[cur].r=update(p[rot].r,mid+1,r,pos);pushup(cur);return cur; } inline int query(int lrot,int rrot,int l,int r,int k) {if(l==r) //到達葉子節點后{if(p[rrot].sum-p[lrot].sum>0) return l;//如果這個值出現過就返回這個值,沒有就返回無窮大else return inf;}int mid=l+r>>1;if(k<=mid){int ans=inf;if(k<=l)//這里相當于一個剪枝,在小于l的時候,如果左子樹有符合條件的就進入左子樹,否則再進入右子樹。{if(p[p[rrot].l].sum-p[p[lrot].l].sum>0) ans=min(ans,query(p[lrot].l,p[rrot].l,l,mid,k));else if(p[p[rrot].r].sum-p[p[lrot].r].sum>0) ans=min(ans,query(p[lrot].r,p[rrot].r,mid+1,r,k));return ans;}if(p[p[rrot].l].sum-p[p[lrot].l].sum>0) //k在l到mid之間的時候,左右子樹都有可能涉及,就左右都看一遍尋找最優解ans=min(ans,query(p[lrot].l,p[rrot].l,l,mid,k));if(p[p[rrot].r].sum-p[p[lrot].r].sum>0) ans=min(ans,query(p[lrot].r,p[rrot].r,mid+1,r,k));return ans;}else{int ans=inf;//k大于mid的時候,直接進入右子樹,左子樹不用找了if(p[p[rrot].r].sum-p[p[lrot].r].sum>0) ans=min(ans,query(p[lrot].r,p[rrot].r,mid+1,r,k));return ans;} }int main() {int t,t1,t2,t3,op;scanf("%d",&t);ll lastans=0;while(t--){scanf("%d%d",&n,&m);set<ll> s;s.insert(n+1);//一開始先把n+1放入集合,如果數組中的值沒有符合的就輸出n+1lastans=0;tot=0;for(int i=1;i<=n;i++) scanf("%d",&a[i]);root[0]=build(1,100001);for(int i=1;i<=n;i++) root[i]=update(root[i-1],1,100001,a[i]);while(m--){scanf("%d",&op);if(op==1){scanf("%d",&t1);t1^=lastans;s.insert(a[t1]);//把修改之后的值放入集合里}else if(op==2){scanf("%d%d",&t2,&t3);t2^=lastans,t3^=lastans;int ans=inf;if(t2!=n) ans=query(root[t2],root[n],1,100001,t3);int ans1=*s.lower_bound(t3);lastans=min(ans1,ans);printf("%d\n",lastans);}}}return 0; }

    昨天寫的超時的代碼(線段樹+set)當個借鑒吧
    代碼如下:

    #include<bits/stdc++.h> #define ll long long #define inf 0x3f3f3f3f using namespace std;const int maxx=1e5+100; struct node{int l;int r;set<int> s; }p[maxx<<2]; int a[maxx]; int n,m;inline int read() //輸入外掛 {int res=0,ch,flag=0;if((ch=getchar())=='-')flag=1;else if(ch>='0'&&ch<='9')res=ch-'0';while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';return flag?-res:res; } inline void Out(int a) //輸出外掛 {if(a>9)Out(a/10);putchar(a%10+'0'); } inline void pushup(int cur) {set_union(p[cur<<1].s.begin(),p[cur<<1].s.end(),p[cur<<1|1].s.begin(),p[cur<<1|1].s.end(),inserter(p[cur].s,p[cur].s.begin())); } inline void build(int l,int r,int cur) {p[cur].l=l;p[cur].r=r;p[cur].s.clear();if(l==r) {p[cur].s.insert(a[l]);return ;}int mid=l+r>>1;build(l,mid,cur<<1);build(mid+1,r,cur<<1|1);pushup(cur); } inline set<int> query(int l,int r,int cur) {int L=p[cur].l;int R=p[cur].r;if(l<=L&&R<=r) return p[cur].s;int mid=L+R>>1;if(r<=mid) return query(l,r,cur<<1);else if(l>mid) return query(l,r,cur<<1|1);else{set<int> a=query(l,mid,cur<<1);set<int> b=query(mid+1,r,cur<<1|1);set<int> x;set_union(a.begin(),a.end(),b.begin(),b.end(),inserter(x,x.begin()));return x;} } int main() {int t,t1,t2,t3,op;t=read();while(t--){int lastans=0;n=read(),m=read();for(int i=1;i<=n;i++) a[i]=read();build(1,n,1);set<int> s;s.insert(n+1);while(m--){op=read();if(op==1){t1=read();t1^=lastans;s.insert(a[t1]);}else{t2=read(),t3=read();t2^=lastans;t3^=lastans;int ans1=inf;if(t2!=n) {set<int> s1;s1=query(t2+1,n,1);s1.insert(n+1);ans1=*s1.lower_bound(t3);}int ans2=*s.lower_bound(t3);lastans=min(ans1,ans2);Out(lastans);}}} }

    努力加油a啊,(o)/~

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