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;
}
/*
*/