Codeforces 887D Ratings and Reality Shows

Ratings and Reality Shows

参加talk show的时间肯定是在某个t[ i ]的后一秒， 枚举一下就好了。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 3e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
const double PI = acos(-1);

int Log[N];
struct ST {
    LL dp[N][20]; int ty;
    void build(int n, LL b[], int _ty) {
        ty = _ty;
        for(int i = -(Log[0]=-1); i < N; i++)
        Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
        for(int i = 1; i <= n; i++) dp[i][0] = ty * b[i];
        for(int j = 1; j <= Log[n]; j++)
            for(int i = 1; i+(1<<j)-1 <= n; i++)
                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    }
    inline LL query(int x, int y) {
        int k = Log[y - x + 1];
        return ty * max(dp[x][k], dp[y-(1<<k)+1][k]);
    }
};

int n, a, b, c, d, start, len;
int t[N], q[N];
LL sum[N];
ST rmq;

int main() {
    cin >> n >> a >> b >> c >> d >> start >> len;
    for(int i = 1; i <= n; i++) {
        cin >> t[i] >> q[i];
        if(q[i]) sum[i] = sum[i - 1] + c;
        else sum[i] = sum[i - 1] - d;
    }
    t[0] = -1;
    rmq.build(n, sum, -1);
    LL prefix = start;
    int ans = -1;
    for(int i = 0; i <= n; i++) {
        if(i) {
            if(q[i]) prefix += a;
            else prefix -= b;
        }
        if(prefix < 0) break;
        int p = lower_bound(t + 1, t + 1 + n, t[i] + 1 + len) - t - 1;
        if(p <= i) {
            ans = t[i] + 1;
            break;
        } else {
            int L = i + 1, R = p;
            LL mn = rmq.query(L, R) - sum[i];
            if(prefix + mn >= 0) {
                ans = t[i] + 1;
                break;
            }
        }
    }
    printf("%d
", ans);
    return 0;
}

/*
*/