解题报告 『[SCOI2010]序列操作（ODT）』

原题地址

用珂朵莉树水题。

代码实现如下：

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

const int maxn = 1e5 + 5;

int n, m;
int a[maxn];

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

set<node> s;

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

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

void rever(int l, int r) {
    IT itr = split(r + 1), itl = split(l);
    while (itl != itr) {
        itl->v ^= 1;
        itl++;
    }
}

int sum(int l, int r) {
    IT itr = split(r + 1), itl = split(l);
    int ans = 0;
    while (itl != itr) {
        ans += itl->v ? itl->r - itl->l + 1 : 0;
        itl++;
    }   
    return ans;
}

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

void write(int x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}

signed main() {
    n = read(), m = read();
    rep(i, 0, n - 1)  {
        a[i] = read();
        s.insert(node(i, i, a[i]));
    }
    s.insert(node(n, n, 0));
    rep(i, 1, m) {
        int l, r, opt;
        opt = read(), l = read(), r = read();
        if(!opt) assign(l, r, 0);
        else if (opt == 1) assign(l, r, 1);
        else if (opt == 2) rever(l, r);
        else if (opt == 3){
            write(sum(l, r));
            printf("\n");
        }
        else {
            write(con_sum(l, r));
            printf("\n");
        }
    }
    return 0;
}

标签：itl,ch,int,ODT,解题,split,ans,itr,SCOI2010
来源： https://www.cnblogs.com/Kirisame-Marisa/p/11364393.html