CF1108E1 Array and Segments (Easy version)（待更新）

题目地址：CF1108E1

加强版题解

我们假设最终的序列中最大的一项为 \(a_x\) ，贪心可得，为了使 \(ans\) 尽量大，我们在操作中不应该对包含 \(x\) 的区间进行操作，而其它区间都应该进行操作

枚举 \(x\) 即可，时间复杂度 \(O(n^2+nm)\)

代码：

#include <bits/stdc++.h>
using namespace std;
const int N = 306;
int n, m, a[N], l[N], r[N], c[N];
vector<int> ans[N];

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= m; i++) scanf("%d %d", &l[i], &r[i]);
    int num = 0, now;
    for (int x = 1; x <= n; x++) {
        memset(c, 0, sizeof(c));
        for (int i = 1; i <= m; i++) {
            if (l[i] <= x && r[i] >= x) continue;
            --c[l[i]];
            ++c[r[i]+1];
            ans[x].push_back(i);
        }
        int s = 0, t = 0, k;
        for (int i = 1; i <= n; i++) {
            s += c[i];
            if (!t || a[i] + s < k) k = a[t=i] + s;
        }
        if (!num || a[x] - k > now) now = a[num=x] - k;
    }
    cout << now << endl << ans[num].size() << endl;
    for (unsigned int i = 0; i < ans[num].size(); i++)
        printf("%d ", ans[num][i]);
    puts("");
    return 0;
}