D. Distinct Characters Queries
Description
You are given a string ss consisting of lowercase Latin letters and qq queries for this string.
Recall that the substring s[l;r]s[l;r] of the string ss is the string slsl+1…srslsl+1…sr. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
There are two types of queries:
- 1 pos c1 pos c (1≤pos≤|s|1≤pos≤|s|, cc is lowercase Latin letter): replace sposspos with cc (set spos:=cspos:=c);
- 2 l r2 l r (1≤l≤r≤|s|1≤l≤r≤|s|): calculate the number of distinct characters in the substring s[l;r]s[l;r].
Input
The first line of the input contains one string ss consisting of no more than 105105 lowercase Latin letters.
The second line of the input contains one integer qq (1≤q≤1051≤q≤105) — the number of queries.
The next qq lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.
output
For each query of the second type print the answer for it — the number of distinct characters in the required substring in this query.
Examples
Input
abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7
Output
3
1
2
正确解法：
原本想着可修改的主席树来着，树状数组加上主席树。
改模板改了好久没过。
操作一：改变某个字符
操作二：统计区间【l，r】不同字符的个数。
26个字母的树状数组：
每次需要for26次，统计这个字母是否在区间内。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <algorithm> 5 #include <set> 6 #include <queue> 7 #include <stack> 8 #include <string> 9 #include <cstring> 10 #include <vector> 11 #include <map> 12 //#include <unordered_map> 13 #define mem( a ,x ) memset( a , x ,sizeof(a) ) 14 #define rep( i ,x ,y ) for( int i = x ; i<=y ;i++ ) 15 #define lson l ,mid ,pos<<1 16 #define rson mid+1 ,r ,pos<<1|1 17 using namespace std; 18 typedef long long ll ; 19 typedef pair<int ,int> pii; 20 typedef pair<ll ,int> pli; 21 const int inf = 0x3f3f3f3f; 22 const int N = 1e5+100; 23 const ll mod =1e9+7 ; 24 char s[N],kkk; 25 int n,m; 26 int aa,bb,cc; 27 int bit[30][N]; 28 int bitsize(int x) 29 { 30 return x&(-x); 31 } 32 void update(int k,int id,int x) 33 { 34 while(id<=n) 35 { 36 bit[k][id]+=x; 37 id+=bitsize(id); 38 } 39 } 40 int query(int k,int id) 41 { 42 int ans=0; 43 while(id>0) 44 { 45 ans+=bit[k][id]; 46 id-=bitsize(id); 47 } 48 return ans; 49 } 50 int main() 51 { 52 scanf("%s",s+1); 53 n=strlen(s+1); 54 for(int i=1;i<=n;i++) 55 { 56 update(s[i]-'a'+1,i,1); 57 } 58 scanf("%d",&m); 59 while(m--) 60 { 61 scanf("%d",&aa); 62 if(aa==1) 63 { 64 scanf("%d %c",&bb,&kkk); 65 update(s[bb]-'a'+1,bb,-1); 66 s[bb]=kkk; 67 update(s[bb]-'a'+1,bb,1); 68 } 69 else 70 { 71 scanf("%d%d",&bb,&cc); 72 int ans=0,res; 73 for(int i=1;i<=26;i++) 74 { 75 res=query(i,cc)-query(i,bb-1); 76 if(res>0) ans++; 77 } 78 printf("%d ",ans); 79 } 80 } 81 82 return 0; 83 }
26个字母的线段树：
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <algorithm> 5 #include <set> 6 #include <queue> 7 #include <stack> 8 #include <string> 9 #include <cstring> 10 #include <vector> 11 #include <map> 12 //#include <unordered_map> 13 #define mem( a ,x ) memset( a , x ,sizeof(a) ) 14 #define rep( i ,x ,y ) for( int i = x ; i<=y ;i++ ) 15 #define lson l ,mid ,pos<<1 16 #define rson mid+1 ,r ,pos<<1|1 17 using namespace std; 18 typedef long long ll ; 19 typedef pair<int ,int> pii; 20 typedef pair<ll ,int> pli; 21 const int inf = 0x3f3f3f3f; 22 const int N = 1e5+100; 23 const ll mod =1e9+7 ; 24 char s[N],kkk; 25 int n,m; 26 int aa,bb,cc; 27 int tree[30][4*N],ans[30],a[N]; 28 void push_up(int rt) 29 { 30 for(int i=1;i<=26;i++) 31 tree[i][rt]=tree[i][rt<<1]+tree[i][rt<<1|1]; 32 } 33 void build(int rt,int l,int r) 34 { 35 if(l==r) 36 { 37 tree[a[l]][rt]++; 38 return; 39 } 40 int mid=l+r >>1; 41 build(rt<<1,l,mid); 42 build(rt<<1|1,mid+1,r); 43 push_up(rt); 44 } 45 void update(int rt,int p,int x,int y,int l,int r) 46 { 47 if(l==r) 48 { 49 tree[x][rt]--; 50 tree[y][rt]++; 51 return; 52 } 53 int mid=l+r>>1; 54 if(p<=mid) 55 update(rt<<1,p,x,y,l,mid); 56 else 57 update(rt<<1|1,p,x,y,mid+1,r); 58 push_up(rt); 59 } 60 void query(int l,int r,int rt,int L,int R) 61 { 62 if(r<=R&&L<=l) 63 { 64 for(int i=1;i<=26;i++) 65 ans[i]+=tree[i][rt]; 66 return ; 67 } 68 int mid=l+r>>1; 69 if(L<=mid) query(l,mid,rt<<1,L,R); 70 if(R>mid) query(mid+1,r,rt<<1|1,L,R); 71 } 72 int main() 73 { 74 scanf("%s",s+1); 75 n=strlen(s+1); 76 for(int i=1;i<=n;i++) 77 a[i]=s[i]-'a'+1; 78 build(1,1,n); 79 scanf("%d",&m); 80 while(m--) 81 { 82 scanf("%d",&aa); 83 if(aa==1) 84 { 85 scanf("%d %c",&bb,&kkk); 86 update(1,bb,s[bb]-'a'+1,kkk-'a'+1,1,n); 87 s[bb]=kkk; 88 } 89 else 90 { 91 scanf("%d%d",&bb,&cc); 92 int sum=0; 93 memset(ans,0,sizeof(ans)); 94 query(1,n,1,bb,cc); 95 for(int i=1;i<=26;i++) 96 if(ans[i]) 97 sum++; 98 printf("%d ",sum); 99 } 100 } 101 102 return 0; 103 }
统计每个字母的下标。用set来快速删除加入元素。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <algorithm> 5 #include <set> 6 #include <queue> 7 #include <stack> 8 #include <string> 9 #include <cstring> 10 #include <vector> 11 #include <map> 12 //#include <unordered_map> 13 #define mem( a ,x ) memset( a , x ,sizeof(a) ) 14 #define rep( i ,x ,y ) for( int i = x ; i<=y ;i++ ) 15 #define lson l ,mid ,pos<<1 16 #define rson mid+1 ,r ,pos<<1|1 17 using namespace std; 18 typedef long long ll ; 19 typedef pair<int ,int> pii; 20 typedef pair<ll ,int> pli; 21 const int inf = 0x3f3f3f3f; 22 const int N = 1e5+100; 23 const ll mod =1e9+7 ; 24 char s[N],kkk; 25 int n,m; 26 int aa,bb,cc; 27 set<int>st[30]; 28 set<int>::iterator it; 29 int main() 30 { 31 scanf("%s",s+1); 32 n=strlen(s+1); 33 for(int i=1;i<=n;i++) 34 st[s[i]-'a'+1].insert(i); 35 scanf("%d",&m); 36 while(m--) 37 { 38 scanf("%d",&aa); 39 if(aa==1) 40 { 41 scanf("%d %c",&bb,&kkk); 42 st[s[bb]-'a'+1].erase(bb); 43 s[bb]=kkk; 44 st[s[bb]-'a'+1].insert(bb); 45 } 46 else 47 { 48 scanf("%d%d",&bb,&cc); 49 int ans=0; 50 for(int i=1;i<=26;i++) 51 { 52 it=st[i].lower_bound(bb); 53 if(it==st[i].end()) continue; 54 if((*it)<=cc) 55 ans++; 56 } 57 printf("%d ",ans); 58 } 59 } 60 61 return 0; 62 }
哦根据状压的思想，把每个字母变成二进制 （1<<x） 判断是否有这个字母就看二进制那个位置上是否为1
判断字母个数就是看二进制上有多少个1。
如果目前没有这个字母，加上就可以了，如果有了，就不变。
| 很好解决了这个问题。
这样一个线段树就完事了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <algorithm> 5 #include <set> 6 #include <queue> 7 #include <stack> 8 #include <string> 9 #include <cstring> 10 #include <vector> 11 #include <map> 12 //#include <unordered_map> 13 #define mem( a ,x ) memset( a , x ,sizeof(a) ) 14 #define rep( i ,x ,y ) for( int i = x ; i<=y ;i++ ) 15 #define lson l ,mid ,pos<<1 16 #define rson mid+1 ,r ,pos<<1|1 17 using namespace std; 18 typedef long long ll ; 19 typedef pair<int ,int> pii; 20 typedef pair<ll ,int> pli; 21 const int inf = 0x3f3f3f3f; 22 const int N = 1e5+100; 23 const ll mod =1e9+7 ; 24 char s[N],kkk; 25 int n,m; 26 int aa,bb,cc; 27 int tree[4*N],ans[30],a[N]; 28 void push_up(int rt) 29 { 30 tree[rt]=tree[rt<<1]|tree[rt<<1|1]; 31 } 32 void build(int rt,int l,int r) 33 { 34 if(l==r) 35 { 36 tree[rt]=(1<<a[l]); 37 return; 38 } 39 int mid=l+r >>1; 40 build(rt<<1,l,mid); 41 build(rt<<1|1,mid+1,r); 42 push_up(rt); 43 } 44 void update(int rt,int p,int x,int l,int r) 45 { 46 if(l==r) 47 { 48 tree[rt]=1<<x; 49 return; 50 } 51 int mid=l+r>>1; 52 if(p<=mid) 53 update(rt<<1,p,x,l,mid); 54 else 55 update(rt<<1|1,p,x,mid+1,r); 56 push_up(rt); 57 } 58 int query(int l,int r,int rt,int L,int R) 59 { 60 if(r<=R&&L<=l) 61 { 62 return tree[rt]; 63 } 64 int mid=l+r>>1,ans=0; 65 if(L<=mid) ans|=query(l,mid,rt<<1,L,R); 66 if(R>mid) ans|=query(mid+1,r,rt<<1|1,L,R); 67 return ans; 68 } 69 int solve(int x) 70 { 71 int ans=0; 72 while(x) 73 { 74 if(x%2==1) ans++; 75 x/=2; 76 } 77 return ans; 78 } 79 int main() 80 { 81 scanf("%s",s+1); 82 n=strlen(s+1); 83 for(int i=1;i<=n;i++) 84 a[i]=s[i]-'a'; 85 build(1,1,n); 86 scanf("%d",&m); 87 while(m--) 88 { 89 scanf("%d",&aa); 90 if(aa==1) 91 { 92 scanf("%d %c",&bb,&kkk); 93 update(1,bb,kkk-'a',1,n); 94 s[bb]=kkk; 95 } 96 else 97 { 98 scanf("%d%d",&bb,&cc); 99 printf("%d ",solve(query(1,n,1,bb,cc))); 100 } 101 } 102 103 return 0; 104 }
