首页 > 代码库 > ZOJ2112--Dynamic Rankings (动态区间第k大)
ZOJ2112--Dynamic Rankings (动态区间第k大)
The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They have developed a more powerful system such that for N numbers a[1], a[2], ..., a[N], you can ask it like: what is the k-th smallest number of a[i], a[i+1], ..., a[j]? (For some i<=j, 0<k<=j+1-i that you have given to it). More powerful, you can even change the value of some a[i], and continue to query, all the same.
Your task is to write a program for this computer, which
- Reads N numbers from the input (1 <= N <= 50,000)
- Processes M instructions of the input (1 <= M <= 10,000). These instructions include querying the k-th smallest number of a[i], a[i+1], ..., a[j] and change some a[i] to t.
Input
The first line of the input is a single number X (0 < X <= 4), the number of the test cases of the input. Then X blocks each represent a single test case.
The first line of each block contains two integers N and M, representing N numbers and M instruction. It is followed by N lines. The (i+1)-th line represents the number a[i]. Then M lines that is in the following format
Q i j k or
C i t
It represents to query the k-th number of a[i], a[i+1], ..., a[j] and change some a[i] to t, respectively. It is guaranteed that at any time of the operation. Any number a[i] is a non-negative integer that is less than 1,000,000,000.
There‘re NO breakline between two continuous test cases.
Output
For each querying operation, output one integer to represent the result. (i.e. the k-th smallest number of a[i], a[i+1],..., a[j])
There‘re NO breakline between two continuous test cases.
Sample Input
2
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
Sample Output
3
6
3
6
主席树动态第k大基本可以手写，，但是感觉理解还不是很深。另外定义两个数组，一个用来新建一颗主席树，所有修改的结果都在这个上面进行，而另一个是记录中间值，相当于temp的效果，不改变原始数组。。（挖个坑）
附主席树专题链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=63941#overview
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <algorithm> 5 #include <iostream> 6 using namespace std; 7 typedef long long ll; 8 const int maxn = 5e4+10; 9 int n,m,tot,idx; 10 ll a[maxn],vec[maxn*2]; 11 struct 12 { 13 int x,y,k,flag,idx; 14 } Q[maxn]; 15 16 // 主席树 17 int lson[maxn*50],rson[maxn*50],c[maxn*50],root[maxn]; //依次为左儿子 右儿子 线段树 根节点 18 int build (int l,int r) 19 { 20 int root = tot++; 21 c[root] = 0; 22 if (l != r) 23 { 24 int mid = (l + r) >> 1; 25 lson[root] = build(l,mid); 26 rson[root] = build(mid+1,r); 27 } 28 return root; 29 } 30 int update(int root,int pos,int val) 31 { 32 int new_root = tot++; 33 int tmp = new_root; 34 int l = 1,r = idx; 35 c[new_root] = c[root] + val; 36 while (l < r) 37 { 38 int mid = (l + r) >> 1; 39 if (pos <= mid) 40 { 41 rson[new_root] = rson[root]; 42 root = lson[root]; 43 lson[new_root] = tot++; 44 new_root = lson[new_root]; 45 r = mid; 46 } 47 else 48 { 49 lson[new_root] = lson[root]; 50 root = rson[root]; 51 rson[new_root] = tot++; 52 new_root = rson[new_root]; 53 l = mid + 1; 54 } 55 c[new_root] = c[root] + val; 56 } 57 return tmp; 58 } 59 // 树状数组维护 60 int s[maxn],use[maxn]; 61 inline int lowbit (int x) 62 { 63 return x & -x; 64 } 65 void add(int k,int pos,int d) 66 { 67 while (k <= n) 68 { 69 s[k] = update(s[k],pos,d); 70 k += lowbit(k); 71 } 72 } 73 int sum(int pos) 74 { 75 int res = 0; 76 while (pos) 77 { 78 res += c[lson[use[pos]]]; 79 pos -= lowbit(pos); 80 } 81 return res; 82 } 83 int query(int left,int right,int k) 84 { 85 int l_root = root[left-1]; 86 int r_root = root[right]; 87 for (int i = left-1; i > 0; i -= lowbit(i)) 88 use[i] = s[i]; 89 for (int i = right; i > 0; i -= lowbit(i)) 90 use[i] =s[i]; 91 int l = 1,r = idx; 92 while (l < r) 93 { 94 int t = sum(right) - sum(left-1) + c[lson[r_root]] - c[lson[l_root]]; 95 int mid = (l + r) >> 1; 96 if (t >= k) 97 { 98 for (int i = left-1; i > 0; i -= lowbit(i)) 99 use[i] = lson[use[i]];100 for (int i = right; i > 0; i -= lowbit(i))101 use[i] = lson[use[i]];102 l_root = lson[l_root];103 r_root = lson[r_root];104 r = mid;105 }106 else107 {108 for (int i = left-1; i > 0; i -= lowbit(i))109 use[i] = rson[use[i]];110 for (int i = right; i > 0; i -= lowbit(i))111 use[i] = rson[use[i]];112 l_root = rson[l_root];113 r_root = rson[r_root];114 k -= t;115 l = mid + 1;116 }117 }118 return l;119 }120 121 void read()122 {123 scanf ("%d%d",&n,&m);124 for (int i = 1; i <= n; i++)125 {126 scanf ("%lld",a+i);127 vec[idx++] = a[i];128 }129 for (int i = 0; i < m; i++)130 {131 char op[3];132 scanf ("%s",op);133 if (op[0] == ‘C‘)134 {135 scanf ("%d%d",&Q[i].x,&Q[i].y);136 Q[i].flag = 0;137 vec[idx++] = Q[i].y;138 }139 if (op[0] == ‘Q‘)140 {141 scanf ("%d%d%d",&Q[i].x,&Q[i].y,&Q[i].k);142 Q[i].flag = 1;143 }144 }145 }146 int main(void)147 {148 #ifndef ONLINE_JUDGE149 freopen("in.txt","r",stdin);150 #endif151 int T;152 scanf ("%d",&T);153 while (T--)154 {155 idx = tot = 0;156 read();157 sort(vec,vec+idx); //离散化158 idx = unique(vec,vec+idx) - vec;159 root[0] = build(1,idx);160 for (int i = 1; i <= n; i++)161 {162 int tmp = lower_bound(vec,vec+idx,a[i]) - vec ;163 root[i] = update(root[i-1],tmp,1);164 }165 for (int i = 0; i <= n; i++)166 s[i] = root[0];167 for (int i = 0; i < m; i++)168 {169 if (Q[i].flag == 0)170 {171 int tmp1 = lower_bound(vec,vec+idx,a[Q[i].x]) - vec ;172 int tmp2 = lower_bound(vec,vec+idx,Q[i].y) - vec ;173 add(Q[i].x,tmp1,-1);174 add(Q[i].x,tmp2,1);175 a[Q[i].x] = Q[i].y;176 }177 if (Q[i].flag == 1)178 {179 printf("%lld\n",vec[query(Q[i].x,Q[i].y,Q[i].k)]);180 }181 }182 }183 return 0;184 }
ZOJ2112--Dynamic Rankings (动态区间第k大)