Cf558e-a Simple Task Segment tree

Give a lowercase letter string, and then Q operation, A,b,c, C is 0, the interval [a, b] by the inverse dictionary sequence, C is 1, this interval is sorted by dictionary order. Using a line segment tree to maintain the number of each letter of each node, after each operation, the violence will be composed of the letter a interval, the letter B, and so on, all elements of the interval is updated to a, B, and so on. Make a lazy mark, and the interval update will be done.

#pragmaComment (linker, "/stack:102400000,102400000")#define_crt_secure_no_warnings#include<stdio.h>#include<iostream>#include<string>#include<queue>#include<stack>#include<list>#include<stdlib.h>#include<algorithm>#include<vector>#include<map>#include<cstring>#include<Set>using namespaceStd;typedefLong LongLL;Const intMoD =1000000007;#defineLson l,mid,rt<<1#defineRson mid+1,r,rt<<1|1Const intMAXN = 1e5 + -;intch[maxn<<2][ -];intCOLOR[MAXN <<2];CharSTR[MAXN];voidUpintRT) {     for(inti =0; I < -; i++) {Ch[rt][i]= Ch[rt <<1][i] + ch[rt <<1|1][i]; }}voidDownintLintRintRT) {    if(Color[rt] = =-1)return; intMid = (L + r) >>1;  for(inti =0; I < -; i++) Ch[rt <<1][i] =0, Ch[rt <<1|1][i] =0; Ch[rt<<1][COLOR[RT]] = (mid-l +1); Ch[rt <<1|1][COLOR[RT]] = R-mid; Color[rt<<1] = Color[rt <<1|1] =Color[rt]; COLOR[RT]= -1;}voidBuildintLintRintRT) {Color[rt]= -1; if(L = =r) {intt = str[l]-'a'; Ch[rt][t]++; return; }    intMid = (L + r) >>1;    Build (Lson);    Build (Rson); Up (RT);}intAskintLintRintAddintLintRintRT) {    if(L <= L&&r <=R) {returnCh[rt][add]; }    intMid = (L + r) >>1; intAns =0;    Down (l, R, RT); if(L <= mid) ans + =Ask (L, R, add, Lson); if(R > mid) ans + =Ask (L, R, add, Rson); returnans;}voidUpdateintLintRintAddintLintRintRT) {    if(L <= L&&r <=R) { for(inti =0; I < -; i++) Ch[rt][i] =0; Ch[rt][add]= (R-l +1); COLOR[RT]=add; return;    } down (l, R, RT); intMid = (L + r) >>1; if(L <=mid) Update (L, R, add, Lson); if(R >mid) Update (L, R, add, Rson); Up (RT);}intAns =0;voidShowintLintRintRT) {    if(L = =r) { for(inti =0; I < -; i++)if(Ch[rt][i]) printf ("%c", i +'a'); return;    } down (L,R,RT); intMid = (L + r) >>1; Show (Lson); Show (Rson);}intcc[ -];intMain () {//freopen ("1.in", "R", stdin); //freopen ("2.in", "w", stdout);    intq, N; intA, B, C; CIN>> N >>Q; scanf ("%s", str); Build (0, N-1,1);  while(q--) {scanf ("%d%d%d", &a, &b, &c); A--; b--; if(c = =1) {             for(inti =0; I < -; i++) Cc[i +1] = Ask (A, B, I,0N1,1); intLL =A;  for(inti =0; I < -; i++) {                if(Cc[i +1] ==0)Continue; Update (LL, LL+ Cc[i +1] -1I0N1,1); ll= ll + cc[i +1]; }        }        Else {             for(inti =0; I < -; i++) Cc[i +1] = Ask (A, B, I,0N1,1); intLL =A;  for(inti = -; I >=0; i--) {                if(Cc[i +1] ==0)Continue; Update (LL, LL+ Cc[i +1] -1I0N1,1); ll= ll + cc[i +1]; }}} Show (0N1,1); return 0;}

Cf558e-a Simple Task Segment tree