【枚举】CF1660 D

Problem - 1660D - Codeforces

题意：

思路：

思路巨简单，代码也wa了很多发才过，都是因为细节....

很显然，要根据0分段处理

对于每一段，枚举去掉左边段还是右边段，左边段是 l 到第一个负数，右边段是最后一个负数到 r，看哪个大

比较的话不需要把区间积算出来，比较区间2的个数即可

如果区间积本来就是正数，那么直接取一整段即可

Code：

#include 

using i64 = long long;

constexpr int N = 2e5 + 10;
constexpr int M = 2e5 + 10;

std::vector > V;

int pl = 1, pr = 1;
int a[N], pre[N], pre2[N];

int calc(int l, int r) {
    if(l == r && a[l] < 0) return -1e9;
    int cnt = pre[r] - pre[l - 1];
    if (cnt & 1) {
        int mx = 0, L = 0, R = 0;
        for (int i = l; i <= r; i ++) {
            if (a[i] < 0) {
                L = i;
                break;
            }
        }

        for (int i = r; i >= l; i --) {
            if (a[i] < 0) {
                R = i;
                break;
            }
        }

        if (pre2[r] - pre2[L + 1 - 1] > pre2[R - 1] - pre2[l - 1]) {
            pl = L + 1;
            pr = r;
        }else {
            pl = l;
            pr = R - 1;
        }
        mx = std::max(pre2[r] - pre2[L + 1 - 1], pre2[R - 1] - pre2[l - 1]);
        return mx;
    }else {
        pl = l, pr = r;
        return pre2[r] - pre2[l - 1];
    }
}
void solve() {
    V.clear();
    pl = 0, pr = 0;
    int n;
    std::cin >> n;

    for (int i = 0; i <= n + 5; i ++) {
        a[i] = pre[i] = pre2[i] = 0;
    }
    for (int i = 1; i <= n; i ++) {
        std::cin >> a[i];
    }

    if (n == 1) {
        if (a[1] > 0) {
            std::cout << "0 0" << "\n";
        }else {
            std::cout << "1 0" << "\n";
        }
        return;
    }
    for (int i = 1; i <= n; i ++) {
        pre[i] = pre[i - 1] + (a[i] < 0);
        pre2[i] = pre2[i - 1] + (abs(a[i]) == 2);
    }
    
    a[n + 1] = 0;
    int l = 1, r = 1;
    for (int i = 1; i <= n + 1; i ++) {
        if (a[i] == 0) {
            r = i - 1;
            if (l <= r) V.push_back({l, r});
            l = i + 1;
        }
    }

    int ans = -1e9, ansl = 0, ansr = 0;

    for (auto v : V) {
        if (ans < calc(v[0], v[1])) {
            ans = calc(v[0], v[1]);
            ansl = pl;
            ansr = pr;
        }
    }

    if (ansl == 0 && ansr == 0) std::cout << n << " " << 0 << "\n";
    else std::cout << ansl - 1 << " " << n - (ansr + 1) + 1 << "\n";
}
signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int t = 1;
    std::cin >> t;
    while(t --) {
        solve();
    }
    return 0;
}