AT_abc195_d [ABC195D] Shipping Center 的题解

AT_abc195_d [ABC195D] Shipping Center 的题解

洛谷传送门

AT传送门

思路

这道题看似很像背包问题，其实不然。我们只需升序排序数组 $X$ 后，再按顺序贪心地为每个盒子选择它能拿到的价值最高的包裹即可。总时间复杂度为 $\mathcal{O}(NMQ)$。

代码

#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <climits>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
#include <ctime>
#include <string>
#include <cstring>
#define lowbit(x) x & (-x)
#define endl "\n"
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
namespace fastIO {
	inline int read() {
		register int x = 0, f = 1;
		register char c = getchar();
		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;
	}
	inline void write(int x) {
		if(x < 0) putchar('-'), x = -x;
		if(x > 9) write(x / 10);
		putchar(x % 10 + '0');
		return;
	}
}
using namespace fastIO;
typedef pair<int, int> pii;
pii bags[55], boxes[55];
bool taken[55];
int main() {
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int n, m, q;
	n = read(), m = read(), q = read();
	for(int i = 0; i < n; i ++) {
		bags[i].second = read(), bags[i].first = read();
	}
	sort(bags, bags + n, greater<pii>());
	for(int i = 0; i < m; i ++){
		boxes[i].first = read();
		boxes[i].second = i;
	}	
	sort(boxes, boxes + m);
	while(q --) {
		int l, r, ans = 0;
		l = read(), r = read();
		l --, r --;
		fill(taken, taken + n, false);
		for(int i = 0; i < m; i ++) {
			auto [size, idx] = boxes[i];
			if(idx < l || idx > r) {
				int j = 0;
				for(; j < n; j ++) {
					if(!taken[j] && bags[j].second <= size) {
						break;
					}
				}
				if(j < n) {
					ans += bags[j].first;
					taken[j] = true;
				}
					
			}
		}
		write(ans), putchar('\n');
	}
	return 0;
}