习题8-23(uva-1623)

最新推荐文章于 2021-07-01 11:56:40 发布

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本文介绍了一段C++代码，用于解决经典的0-1背包问题。通过输入整数数组和对应价值，程序实现了选择物品以最大化总价值，同时不超过背包容量。关键步骤包括使用优先队列和迭代更新策略。

#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int maxn = 1e6 + 7;
int last[maxn],input[maxn],output[maxn],n,m;
int main()
{
	int t;
	cin >> t;
	set<int> mset;
	while (t--) {
		mset.clear();
		cin >> n >> m;
		memset(last, 0, sizeof(last));
		for (int i = 1; i <= m; i++) {
			cin >> input[i];
			output[i] = 0;
		}
		bool ok = true;
		for (int i = 1; i <= m; i++) {
			if (input[i] == 0) {
				mset.insert(i);
			}
			else {
				auto it = mset.lower_bound(last[input[i]]);
				if (it == mset.end()) { ok = false; break; }
				output[*it] = input[i];
				last[input[i]] = i;
				mset.erase(*it);
			}
		}
		if (ok) {
			cout << "YES" << endl;
			int index = 1;
			while (index <= m && input[index] != 0) { ++index; continue; }
			cout << output[index++];
			for (; index <= m; index++) {
				if(input[index] != 0) continue;
				cout <<" "<< output[index];
			}
			cout << endl;
		}
		else cout << "NO" << endl;
	}
	return 0;
}