解题报告 『矿洞：坍塌（ODT）』

原题地址

退役第二天也要坚持用珂朵莉树水题。

 

代码实现如下：

#include <bits/stdc++.h>
using namespace std;
#define IT set<node>::iterator
#define rep(i, a, b) for (register int i = (a); i <= (b); i++)

const int maxn = 5e5 + 5;

int n, m, cnt = 1;
char str[maxn];

struct node {
    int l, r;
    mutable char v;
    node(int L, int R = -1, char V = 0): l(L), r(R), v(V) {}
    int operator <(const node &o) const {
        return l < o.l;
    }
};

set<node> s;

int read() {
    int x = 0, flag = 0;
    char ch = ' ';
    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') {
        flag = 1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + (ch ^ '0');
        ch = getchar();
    }
    return flag ? -x : x;
}

IT split(int pos) {
    IT it = s.lower_bound(node(pos));
    if (it != s.end() && it->l == pos) return it;
    it--;
    int L = it->l, R = it->r;
    char V = it->v;
    s.erase(it);
    s.insert(node(L, pos - 1, V));
    return s.insert(node(pos, R, V)).first;
}

void assign(int l, int r, char val) {
    IT itr = split(r + 1), itl = split(l);
    s.erase(itl, itr);
    s.insert(node(l, r, val));
}

char get_val(int pos) {
    IT it = split(pos);
    return it->v;
}

int check(int l, int r) {
    IT itr = split(r + 1), itl = split(l);
    char now = itl->v;
    while (itl != itr) {
        if (now != itl->v) return 0;
        itl++;
    }
    return 1;
}

int main() {
    n = read();
    scanf("%s", str + 1);
    char pre = str[1];
    rep(i, 2, n) {
        if (pre == str[i]) cnt++;
        else {
            s.insert(node(i - cnt, i - 1, pre));
            cnt = 1;
            pre = str[i];
        }
    }
    s.insert(node(n - cnt + 1, n, pre));
    m = read();
    char tmp[3];
    rep(i, 1, m) {
        int l, r;
        scanf("%s", tmp + 1), l = read(), r = read();
        if (tmp[1] == 'A') {
            char opt[3];
            scanf("%s", opt + 1);
            assign(l, r, opt[1]);
        }
        else {
            if (l == 1 || l == n || r == 1 || r == n) {
                if (check(l, r)) printf("Yes\n");
                else printf("No\n");
                continue;
            }
            char a = get_val(l - 1), b = get_val(r + 1);
            if (a == b || !check(l, r)) printf("No\n");
            else printf("Yes\n");
        }
    }
    return 0;
}

 

ODT并不是这道题的正解，所以不开O2的话只有90pts（当然也有可能是我太蒻了）。

线段树的做法有时间再写吧。