string 26线段树

10.4

思路：
用线段树维护区间内a~z 的个数，每次询问拆成26 个区间修改操作。也就是说处理出一个区间a~z 的个数，然后按升序降序区间修改。常数有点大，需要卡卡。比如query的时候返回一个结构体什么的。

#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>using namespace std;const int maxn = 100010;int a[maxn], n, T, cnt[50], sum[50];char s[maxn];inline const int read(){    register int f = 1, x = 0;    register char ch = getchar();    while(ch < '0' || ch > '9') ch = getchar();    while(ch >= '0' && ch <= '9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();    return x;}struct Node{    Node *ls, *rs;    int cnt[27], flag;    Node(){ flag = 0 ; memset( cnt , 0 , sizeof( cnt )) ;}    Node operator+( const Node &a ) const{        Node tmp = Node();        for(int i=1; i<=26; ++i)            tmp.cnt[i] = cnt[i] + a.cnt[i] ;        return tmp;    }    void pushdown(int lf, int rg){        if( flag ){            int mid = (lf + rg) >> 1;            ls->cnt[flag] = mid - lf + 1;            ls->flag = flag;            rs->cnt[flag] = rg - mid;            rs->flag= flag;            for(int i=1; i<=26; ++i)                 if(i != flag) ls->cnt[i] = rs->cnt[i] = 0;            flag = 0;        }    }    void update(){        for(int i=1; i<=26; ++i)            cnt[i] = ls->cnt[i] + rs->cnt[i];    }}pool[maxn*20], *tail = pool, *rot;inline Node* build(int lf, int rg){    Node *nd = ++tail;    if(lf == rg) {        for(int i=1; i<=26; ++i) nd->cnt[i] = 0;        nd->cnt[a[lf]]++;         return nd;    }    int mid = (lf + rg) >> 1;    nd->ls = build(lf, mid);    nd->rs = build(mid+1, rg);    nd->update();    return nd;}inline void modify(Node *nd, int lf, int rg, int L, int R, int data){    if(L <= lf && rg <= R){        nd->cnt[data] = rg - lf + 1;         nd->flag = data;        for(int i=1; i<=26; ++i)             if(i != data) nd->cnt[i] = 0;        return;    }     nd->pushdown(lf, rg);    int mid = (lf + rg) >> 1;    if(L <= mid) modify(nd->ls, lf, mid, L, R, data);    if(R > mid) modify(nd->rs, mid+1, rg, L, R, data);    nd->update();    return;}inline Node query(Node *nd, int lf, int rg , int L , int R ){    if( L <= lf && rg <= R )        return *nd;    Node rt = Node() ;    nd->pushdown(lf, rg) ;    int mid = ( lf + rg ) >> 1 ;    if( L <= mid ) rt = rt + query( nd->ls, lf , mid , L , R ) ;    if( R >  mid ) rt = rt + query( nd->rs, mid+1, rg , L , R ) ;    return rt ;}//返回一个结构体（内有cnt[1~26]），就只需要run一次，而不是26次了。 inline int query_point(Node *nd, int lf, int rg, int pos){    if(lf == rg){        for(int i=1; i<=26; ++i)            if(nd->cnt[i]) return i;    }    nd->pushdown(lf, rg);    int mid = (lf + rg) >> 1, rt = 0;    if(pos <= mid) rt = query_point(nd->ls, lf, mid, pos);    else rt = query_point(nd->rs, mid+1, rg, pos);    nd->update();    return rt;}int main(){    freopen("string.in", "r", stdin);    freopen("string.out", "w", stdout);    n = read(), T = read();    scanf("%s", s+1);    for(register int i=1; i<=n; ++i) a[i] = s[i] - 'a' + 1;    rot = build(1, n);    for(int i=1; i<=T; ++i){        int l = read(), r = read(), opt = read();        Node t = query(rot, 1, n, l, r);//答案结构体         for(int j=1; j<=26; ++j) sum[j] = t.cnt[j];        int st = l;        if( opt ){            for(int j=1; j<=26; ++j)                if( sum[j] ) modify(rot, 1, n, st, st+sum[j]-1, j), st += sum[j];        }        else{            for(int j=26; j>=1; --j)                if( sum[j] ) modify(rot, 1, n, st, st+sum[j]-1, j), st += sum[j];        }    }    for(register int j=1; j<=n; ++j) printf("%c", query_point(rot, 1, n, j) + 'a' - 1);    return 0;}/*5 2cabcd1 3 13 5 0*/