洛谷P4064 [JXOI2017]加法(貪心 差分)

題意

題目鏈接

Sol

這題就是一個很顯然的貪心。。。

首先二分一個答案，然後check是否可行。check的時候我們需要對每個位置\(i\)，維護出所有左端點在\(i\)左側，右端點在\(i\)右側的所有區間。最優策略一定是加右端點最遠的。

然後就做完了， 複雜度\(O(nlogn)\)

#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;
}
/*
*/