洛谷P4064 [JXOI2017]加法(贪心 差分)

题意

题目链接

sol

这题就是一个很显然的贪心。。。

首先二分一个答案，然后check是否可行。check的时候我们需要对每个位置\(i\)，维护出所有左端点在\(i\)左侧，右端点在\(i\)右侧的所有区间。最优策略一定是加右端点最远的。

然后就做完了， 复杂度\(o(nlogn)\)

#include<bits/stdc++.h>
#define fin(x) freopen(#x".in", "r", stdin);
#define ll long long 
#define int long long 
using namespace std;
const int maxn = 1e5 + 10;
const ll inf = 1e18;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m, k, a;
vector<int> v[maxn];
ll a[maxn], b[maxn];
bool check(int mid) {
    memset(b, 0, sizeof(b));
    priority_queue<int> q;
    int tag = 0, num = 0;
    for(int i = 1; i <= n; i++) {
        for(auto &x : v[i]) q.push(x);
        while(!q.empty() && q.top() < i) q.pop();
        tag += b[i];
        int now = a[i] + tag;
        while(now < mid && !q.empty() && q.top() >= i) {
            b[q.top() + 1] -= a; tag += a; q.pop();
            now += a; num++;
        }
        if(now < mid || num > k) return 0;
    }
    if(num <= k) return 1;
}
void solve() {
    n = read(); m = read(); k = read(); a = read();
    for(int i = 1; i <= n; i++) a[i] = read(), v[i].clear();
    while(m--) {
        int l = read(), r = read();
        v[l].push_back(r);
    }
    int l = 0, r = inf, ans = -1;
    while(l <= r) {
        int mid = l + r >> 1;
        if(check(mid)) l = mid + 1, ans = mid;
        else r = mid - 1;
    }
    cout << ans << '\n';
}
signed main() {
    for(int t = read(); t--; solve());
    return 0;
}
/*
*/