Codeforces 899F Letters Removing 线段树/树状数组

  虽然每次给一个区间，但是可以看作在区间内进行数个点操作，同样数列下标是动态变化的，如果我们将每个字符出现看作1，被删除看作0，则通过统计前缀和就能轻松计算出两个端点的位置了！这正是经典的树状数组操作

需要注意的是，即使每次找到了两个端点，如果只是朴素总左到右遍历会TLE，所以可以给每个字符开一个set，每个set储存字符出现的位置。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
#pragma warning ( disable : 4996 )
using namespace std;

int Max( int a, int b ) { return a>b?a:b; }
int Min( int a, int b ) { return a>b?b:a; }
int lowbit( int x )        { return x&-x; }

const int inf = 0x3f3f3f3f;
const int maxn = 2e5+5;

set<int> pos[65];
set<int>::iterator it, it2;
char str[maxn];
int istr[maxn];
bool isdel[maxn];

int len, q;

int getI( char c )
{
    if( c >= 'a' && c <= 'z' ) return c-'a';
    if( c >= 'A' && c <= 'Z' ) return c-'A'+26;
    if( c >= '0' && c <= '9' ) return c-'0'+52;
}

int sum( int x )
{
    int ret = 0;
    while( x > 0 )
        {    ret += istr[x]; x -= lowbit(x); }
    return ret;
}

void add( int x, int d )
{
    while( x <= len )
        { istr[x] += d; x += lowbit(x); }
}

int find( int s )
{
    int mid, lhs = 1, rhs = len;
    int tmp;
    while ( lhs <= rhs )
    {
        mid = (lhs+rhs)>>1;
        tmp = sum(mid);
        if ( tmp >= s ) rhs = mid-1;
        else lhs = mid+1;
    }
    return lhs;
}

void change( int l, int r, char c )
{
    int p = getI(c);
    it = pos[p].lower_bound(l);
    while ( it!=pos[p].end() && *it<=r )
    {
        isdel[(*it)] = true;
        add((*it), -1);
        it2 = it; it++;
        pos[p].erase(it2);
    }
}

int main()
{
    cin >> len >> q;
    scanf("%s", str+1);

    for ( int i = 1; i <= len; i++ )
    {
        add(i,1);
        int p = getI(str[i]);
        pos[p].insert(i);
    }

    int x, y;
    char c;
    for ( int i = 1; i <= q; i++ )
    {
        scanf( "%d %d %c", &x, &y, &c );
        x = find(x); y = find(y);
        change( x, y, c );
    }

    for( int i = 1; i <= len; i++ )
        if( !isdel[i] )
            printf( "%c", str[i] );
    
    printf("
");
    return 0;
}

 我们看到区间操作，自然就会想到线段树，但是数列下标是动态变换的，乍一看似乎线段树就没有用武之地了。实际仔细想想，线段树也可以动态统计从哪个区间开始操作，相比普通线段树只是在每次操作前要计算新的下标到哪了，实际上每个节点管辖的范围是始终不变的，变的只是范围内字符的个数size，每次操作前通过这个size找到对应的左端点和右端点就行了。

//线段树
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <vector>
#include <algorithm>
#pragma warning ( disable : 4996 )
using namespace std;

int Max( int a, int b ) { return a>b?a:b; }
int Min( int a, int b ) { return a>b?b:a; }

const int inf = 0x3f3f3f3f;
const int maxn = 2e5+2;

struct tree {
    int size, lhs, rhs;
    int clu[62];
}t[maxn<<2];
char str[maxn];

int getI( char c )
{
    if( c >= 'a' && c <= 'z' ) return c-'a';
    if( c >= 'A' && c <= 'Z' ) return c-'A'+26;
    if( c >= '0' && c <= '9' ) return c-'0'+52;
}

char getC( int i )
{
    if( i >= 0 && i <= 25 )  return 'a'+i;
    if( i >= 26 && i <= 51 ) return 'A'+i-26;
    if( i >= 52 && i <= 61 ) return '0'+i-52;
}

void pushUp( int index )
{
    t[index].size = t[index<<1].size + t[index<<1|1].size;
    for ( int i = 0; i < 62; i++ )
        t[index].clu[i] = t[index<<1].clu[i] + t[index<<1|1].clu[i];
}

void build( int l, int r, int index )
{
    t[index].lhs = l; t[index].rhs = r;
    if( l == r )
        { t[index].size = 1; t[index].clu[getI(str[l-1])] = 1; return; }
    int mid = (l+r)>>1;
    build(l, mid, index<<1);
    build(mid+1, r, index<<1|1);
    pushUp(index);
}

int find( int index, int x )
{
    int l = t[index].lhs, r = t[index].rhs;
    if( l == r ) return l;
    if( t[index<<1].size >= x ) return find( index<<1, x );
    else return find( index<<1|1, x-t[index<<1].size );
}

void change( int index, int l, int r, int x ) 
{
    int L = t[index].lhs, R = t[index].rhs;
    if ( r < L || l > R ) return;
    if ( !t[index].clu[x] ) return;
    if ( L == R ) { t[index].size--; t[index].clu[x]--; return; }
    change( index<<1, l, r, x );
    change( index<<1|1, l, r, x );
    t[index].size = t[index<<1].size + t[index<<1|1].size;
    t[index].clu[x] = t[index<<1].clu[x] + t[index<<1|1].clu[x];
}

void print( int i )
{
    int L=t[i].lhs,R=t[i].rhs;
    if ( L==R )
    {
        for ( int j = 0; j < 62; j++ )
            if ( t[i].clu[j] ) printf("%c",getC(j));
        return;
    }
    print(i<<1);
    print(i<<1|1);

}

int main()
{
    int len, q;
    cin >> len >> q;
    scanf("%s", str);
    build(1, len, 1);

    int x, y;
    char c;
    for ( int i = 1; i <= q; i++ )
    {
        scanf( "%d %d %c", &x, &y, &c );
        int l = find( 1, x );
        int r = find( 1, y );
        change( 1, l, r, getI(c) );
    }
    print(1);
    printf("
");
    return 0;
}