HOJ 2660 Accepted Necklace

本题直接使用一个快排和深搜dfs可以解决

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int value_M, tolerance, n, need;

//使用结构体进行存储，方便使用快速排序
typedef struct stone {
	int value;
	int weight;
}Stone;
Stone a[20];

//以宝石weight为首要标准，从大到小排，weight相同情况下value从大到小
int cmp(Stone x, Stone y) {
	if (x.weight == y.weight) return x.value > y.value;
	return x.weight > y.weight;
}

void dfs(int k, int &times, int &value, int &weight) {
	if (times == need) {//所需宝石数量达到目的
		if (value > value_M)//value最大值更新
			value_M = value;
		return;
	}
	for (int i = k + 1; i < n; i++) {
		if (weight + a[i].weight <= tolerance) {//weight要在可接受范围内才可以进行操作
			value += a[i].value;
			weight += a[i].weight;
			times++;
			dfs(i, times, value, weight);//当前宝石符合要求，继续寻找下一个
			value -= a[i].value;
			weight -= a[i].weight;
			times--;
		}
	}
}

int main() {
	int t, value, weight, times;
	while (cin >> t) 
		while (t--) {
			for (int i = 0; i < 20; i++) {//初始化
				a[i].value = 0;
				a[i].weight = 0;
			}
			value = 0;
			weight = 0;
			times = 0;
			value_M = 0;
			cin >> n >> need;
			for (int i = 0; i < n; i++)
				cin >> a[i].value >> a[i].weight;
			cin >> tolerance;
			sort(a, a + n, cmp);
			while (a[0].weight > tolerance) {//以经排好序，把单个weight大于tolerance的宝石丢掉
				for (int i = 0; i < n - 1; i++)
					a[i] = a[i + 1];
				n--;
			}
			dfs(0, times, value, weight);
			cout << value_M << endl;
		}
	return 0;
}