CF2002E

题目链接

手搓一下第一个测试样例可以发现， \(1\sim 3\) 的 \(1,3\) 合并，\(a_3>a_2\) 所以可以维护一个单调栈，单调递减，如果单调栈顶的下一位 \(b\) 等于当前的 \(b\)，且当前 \(a\) 大于栈顶 \(a\) 那么将栈顶下一位和当前的序列合并，这样答案就是栈底。

\(\mathscr{Code:}\)

#include<bits/stdc++.h>
#define LL long long
#define int LL
#define per(i, a, b) for (int i = a, END##i = b; i >= END##i; i--)
#define rep(i, a, b) for (int i = a, END##i = b; i <= END##i; i++)
#define repn(x) rep(x, 1, n)
#define repm(x) rep(x, 1, m)
#define pb push_back
#define e(x) for(int i = h[x], v = to[i]; i; i = nxt[i], v = to[i])
#define E(x) for(auto y : p[x])
#define PII pair<int, int>
#define i64 unsigned long long
#define YY puts("Yes"), exit(0)
#define NN puts("No"), exit(0)
using namespace std;
const int Mod = 1e9 + 7;
const int Inf = 0x3f3f3f3f;
const LL InfLL = 0x3f3f3f3f3f3f3f3f;
inline LL read() {LL s = 0, fu = 1; char ch = getchar(); while (ch < '0' || ch > '9') ch == '-' ? fu = -1 : 0, ch = getchar(); while (ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * fu;}

#define x first
#define y second
const int N = 3e5 + 10;
int n, top;
PII a[N], stk[N];

inline void Main() {
    top = 0;
    n = read();
    repn(i) {
        a[i].x = read();
        a[i].y = read();
    }
    stk[0] = {0, -1000};
    rep(i, 1, n) {
        while (top && stk[top].x <= a[i].x) {
            if (stk[top - 1].y == a[i].y) {
                a[i].x += stk[top - 1].x - stk[top].x;
                top--;
            }
            top--;
        }
        stk[++top] = a[i];
        printf("%lld ", stk[1].x);
    }
    puts("");
}

signed main() {
    // freopen("input.in", "r", stdin);
    int T = read();
    while (T--)
        Main();
    return 0;
}

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