題目鏈接 A Simple Task

題意 給出一個小寫字母序列和若幹操作。每個操作為對給定區間進行升序排序或降序排序。

考慮權值線段樹。

建立26棵權值線段樹。每次操作的時候先把26棵線段樹上的所有在該區間內的信息清空。

然後再通過類似計數排序的方式從左往右（或從右往左）依次塞進去。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define ls		i << 1
#define	rs		i << 1 | 1
#define lson		ls, L, mid
#define	rson		rs, mid + 1, R



typedef long long LL;

const int N = 1e5 + 10;

struct node{
	int sum[2];
	int len;
} s[27][N * 3];

int n, q;
int a[N];
char st[N];
int dc[27][N * 3];
int f[27];


inline node update(const node &x, const node &y){
	node ret;
	ret.len = x.len + y.len;
	rep(i, 0, 1) ret.sum[i] = x.sum[i] + y.sum[i];
	return ret;
}

void build(int cnt, int i, int L, int R){
	if (L == R){
		int z = a[L] == cnt;
		s[cnt][i].len = 1;
		s[cnt][i].sum[z] = 1;
		s[cnt][i].sum[z ^ 1] = 0;
		dc[cnt][i] = -1;
		return;
	}

	int mid = (L + R) >> 1;
	build(cnt, lson);
	build(cnt, rson);
	s[cnt][i] = update(s[cnt][ls], s[cnt][rs]);
	dc[cnt][i] = -1;
}

inline void paintcover(int cnt, int i, int z){
	dc[cnt][i] = z;
	s[cnt][i].sum[z] = s[cnt][i].len;
	s[cnt][i].sum[z ^ 1] = 0;
}

inline void pushdown(int cnt, int i){
	if (~dc[cnt][i]){
		paintcover(cnt, ls, dc[cnt][i]);
		paintcover(cnt, rs, dc[cnt][i]);
		dc[cnt][i] = -1;
	}
}

void cover(int cnt, int i, int L, int R, int l, int r, int z){
	if (l <= L && R <= r){
		paintcover(cnt, i, z);
		return;
	}

	pushdown(cnt, i);
	int mid = (L + R) >> 1;
	if (l <= mid) cover(cnt, lson, l, r, z);
	if (r >  mid) cover(cnt, rson, l, r, z);
	s[cnt][i] = update(s[cnt][ls], s[cnt][rs]);
}

int query(int cnt, int i, int L, int R, int l, int r){
	int ret = 0;
	if (l <= L && R <= r) return s[cnt][i].sum[1];
	int mid = (L + R) >> 1;
	pushdown(cnt, i);
	if (l <= mid) ret += query(cnt, lson, l, r);
	if (r >  mid) ret += query(cnt, rson, l, r);
	return ret;
}

int main(){

	scanf("%d%d", &n, &q);
	scanf("%s", st + 1);
	rep(i, 1, n) a[i] = (int)st[i] - 96;
	rep(i, 1, 26) build(i, 1, 1, n);

	while (q--){
		int x, y, z;
		scanf("%d%d%d", &x, &y, &z);
		rep(i, 1, 26) f[i] = query(i, 1, 1, n, x, y);
		rep(i, 1, 26) cover(i, 1, 1, n, x, y, 0);
		if (z){
			int pos = x;
			rep(i, 1, 26) if (f[i]){
				cover(i, 1, 1, n, pos, pos + f[i] - 1, 1);
				pos += f[i];
			}
		}

		else{
			int pos = x;
			dec(i, 26, 1) if (f[i]){
				cover(i, 1, 1, n, pos, pos + f[i] - 1, 1);
				pos += f[i];
			}
		}
	}

	rep(i, 1, n){
		rep(j, 1, 26) if (query(j, 1, 1, n, i, i)){
			putchar(j + 96);
			break;
		}
	}

	putchar(10);
	return 0;
}

Codeforces 558E A Simple Task（權值線段樹）