A simple simulation problem.（线段树）

题目描述：

There are n types of cells in the lab, numbered from 1 to n. These cells are put in a queue, the i-th cell belongs to type i. Each time I can use mitogen to double the cells in the interval [l, r]. For instance, the original queue is {1 2 3 3 4 5}, after using a mitogen in the interval [2, 5] the queue will be {1 2 2 3 3 3 3 4 4 5}. After some operations this queue could become very long, and I can’t figure out maximum count of cells of same type. Could you help me?

InputThe first line contains a single integer t (1 <= t <= 20), the number of test cases.

For each case, the first line contains 2 integers (1 <= n,m<= 50000) indicating the number of cell types and the number of operations.

For the following m lines, each line represents an operation. There are only two kinds of operations: Q and D. And the format is:

“Q l r”, query the maximum number of cells of same type in the interval [l, r];
“D l r”, double the cells in the interval [l, r];

(0 <= r – l <= 10^8, 1 <= l, r <= the number of all the cells)OutputFor each case, output the case number as shown. Then for each query "Q l r", print the maximum number of cells of same type in the interval [l, r].

Take the sample output for more details.

Solution

处理一个cnt[i]代表颜色i出现的次数

维护一棵线段树，维护区间和。

sum[i]代表前缀和
查询：[a,b]
先找到a的颜色和b的颜色再分类讨论
设其为i,j
1，如果i=j同色，则输出b-a+1;
2，如果j=i+1，则输出max(sum[i]-a+1,b-sum[i])
3，如果j >= i+2，则输出max(max(sum[i]-a+1,b-sum[j-1]),max{cnt[k]})(i<k<j)
如何找到a的颜色
设我们要找x的颜色
在线段树上二分，如果左孩子的和>=x，则递归搜索左孩子
如果小于则地递归搜索右孩子。
时间复杂度:O(N logN)
如何区间修改呢？
和查询类似，先找到a的颜色和b的颜色
设其为i，j
1，如果i=j，则直接单点修改cnt[i]加上j-i+1
2，如果i=j+1，则做两次单点修改cnt[i]加上sum[i]-a+1，cnt[j]加上b-sum[i];
注意这里sum[i]不能分两次求（应该也没有傻子分两次求吧）
3，如果j >= i+2，则对于i，j的操作同操作2，
之后我们对区间(i,j)乘一个2即可
实现就很简单了。

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
const ll N = 50010;
struct Segment{
    ll sum, mx;
    ll mul;
}tr[N << 2];
ll t, n, m;
void push_up(ll p) {
    tr[p].sum = tr[p << 1].sum + tr[p << 1 | 1].sum;
    tr[p].mx = max(tr[p << 1].mx, tr[p << 1 | 1].mx);
}
void push_down(ll p) {
    if (tr[p].mul > 1) {
        tr[p << 1].sum *= tr[p].mul;
        tr[p << 1 | 1].sum *= tr[p].mul;
        tr[p << 1].mx *= tr[p].mul;
        tr[p << 1 | 1].mx *= tr[p].mul;
        tr[p << 1].mul *= tr[p].mul;
        tr[p << 1 | 1].mul *= tr[p].mul;
        tr[p].mul = 1;
    }
}
void build (ll p, ll l, ll r) {
    tr[p].mul = 1;
    if (l == r) {
        tr[p].sum = tr[p].mx = 1;
        return;
    }
    ll mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    push_up(p);
}
//查询颜色 
ll ask_color(ll p, ll l, ll r, ll k) {
//    cout << l << " " << r << endl;
    if (l == r) return l;
    push_down(p);
    ll num = tr[p << 1].sum;
//    cout << num << endl;
    ll mid = (l + r) >> 1;
    if (k <= num) return ask_color(p << 1, l, mid, k);
    else return ask_color(p << 1 | 1, mid + 1, r, k - num);
}
//查询和 
ll ask_sum(ll p, ll l, ll r, ll x, ll y) {
    if (r < x || l > y) return 0;
    if (x <= l && r <= y) {
        return tr[p].sum;
    }
    push_down(p);
    ll mid = (l + r) >> 1;
    return ask_sum(p << 1, l, mid, x, y) + ask_sum(p << 1 | 1, mid + 1, r, x, y);
}
//查询最值 
ll ask_max(ll p, ll l, ll r, ll x, ll y) {
    if (r < x || l > y) return 0;
    if (x <= l && r <= y) {
        return tr[p].mx;
    }
    push_down(p);
    ll mid = (l + r) >> 1;
    return max(ask_max(p << 1, l, mid, x, y), ask_max(p << 1 | 1, mid + 1, r, x, y));
}
//单点修改（+v） 
void add(ll p, ll l, ll r, ll pos, ll v) {
    if (l == r) {
        tr[p].sum += v;
        tr[p].mx += v;
        return;
    }
    push_down(p);
    ll mid = (l + r) >> 1;
    if (pos <= mid) add(p << 1, l, mid, pos, v);
    else add(p << 1 | 1, mid + 1, r, pos, v);
    push_up(p);
}
//区间修改（*2） 
void multi(ll p, ll l, ll r, ll x, ll y) {
    if (l > y || r < x) return;
    if (x <= l && r <= y) {
        tr[p].mul *= 2;
        tr[p].sum *= 2;
        tr[p].mx *= 2;
        return;
    }
    push_down(p);
    ll mid = (l + r) >> 1;
    multi(p << 1, l, mid, x, y);
    multi(p << 1 | 1, mid + 1, r, x, y);
    push_up(p);
}
void change(ll l, ll r) {
    ll i = ask_color(1, 1, n, l);
    ll j = ask_color(1, 1, n, r);
//    cout << i << " " << j << endl;
    if (i == j) {
        add(1, 1, n, i, r - l + 1);
    } else {
        ll sm = ask_sum(1, 1, n, 1, i);
        ll sm2 = ask_sum(1, 1, n, 1, j - 1);
        if (i + 1 == j) {
            add(1, 1, n, i, sm - l + 1);
            add(1, 1, n, j, r - sm); 
        } else {
//            cout << sm << endl;
            add(1, 1, n, i, sm - l + 1);
            ask_color(1, 1, n, l);
//            cout << r - sm2 << endl;
            add(1, 1, n, j, r - sm2);
            ask_color(1, 1, n, l);
            multi(1, 1, n, i + 1, j - 1);
        }
    }
}
ll query(ll l, ll r) {
    ll i = ask_color(1, 1, n, l);
    ll j = ask_color(1, 1, n, r);
//    cout << i << " " << j << endl;
    if (i == j) {
        return r - l + 1;
    } else {
        ll sm = ask_sum(1, 1, n, 1, i);
        ll sm2 = ask_sum(1, 1, n, 1, j - 1);
        if (i + 1 == j) {
            return max(sm - l + 1, r - sm);
        } else {
//            cout << ask_sum(1, 1, n, 1, j - 1) << endl;
//            ll ZSY = max(sm - l + 1, r - ask_sum(1, 1, n, 1, j - 1));
//            cout << ZSY << endl;
            return max(max(sm - l + 1, r - sm2), ask_max(1, 1, n, i + 1, j - 1));
        }
    }
}
ll CRZzzz;
int main() {
    scanf("%lld", &t);
    while (t--) {
        CRZzzz++; 
        scanf("%lld%lld", &n, &m);
        printf("Case #%lld:
", CRZzzz);
        build(1, 1, n);
        while (m--) {
            char s[5];
            ll a, b;
            scanf("%s%lld%lld", s + 1, &a, &b);
            if (s[1] == 'D') {
                change(a, b);
            } else {
                printf("%lld
", query(a, b));
            }
        }
    }
    return 0;
}