UVa - 1623 - Enter The Dragon

The capital of Ardenia is surrounded by several lakes, and each of them is initially full of water. Currently, heavy rainfalls are expected over the land. Such a rain falls to one of the lakes: if the lake is dry
and empty, then it will be filled with water; if the lake is already full, then it will overflow, which will result in a natural disaster. Fortunately, the citizens have a dragon at their disposal (and they will not hesitate to use it). The dragon may drink
the whole water from a lake in one sitting. Also, the mages of Ardenia already predicted the weather conditions for the next couple of years. The only question is: from which lake and when should the dragon drink to prevent a catastrophe?

Input

The input contains several test cases. The first line of the input contains a positive integer Z40,
denoting the number of test cases. Then Z test cases follow, each conforming to the format described below.

The first line of the input instance contains two space-separated positive integers n10⁶ and m10⁶ ,
where n is the number of lakes. (There are at most 10 input instances for which n10⁵ or m10⁵.)
The second line contains the weather forecast for the next m days: m space-separated integers t₁, t₂,..., t_m (t_i  [0, n]).
If t_i  [1, n], it means a heavy rainfall over lake t_i at
day i. If t_i = 0, there is no rain at day i, and the dragon has the time to drink the water from one lake of your choice. Note that the dragon
does not drink on a rainy day.

Output

For each test case, your program has to write an output conforming to the format described below.

In the first line your program should output word `YES‘ if it is possible to prevent a catastrophic overflow and `NO‘ otherwise. In the former case, you should output the second line containing l integers
from the range [0, n], where l is the number of zeros in the weather forecast description, i.e., the number of non-rainy days. Each of these integers denotes the number of the lake from
which the dragon should drink; zero means the dragon should not drink from any lake (this might be necessary, as even the dragon cannot drink from an empty lake).

Sample Input

4
2 4
0 0 1 1
2 4
0 1 0 2
2 3
0 1 2
2 4
0 0 0 1

Sample Output

NO
YES
1 2
NO
YES
0 1 0

贪心，用数组记录每个湖上次满水的日子，用集合记录不下雨的日子。下雨的时候，查找当前湖最后灌满的日子之后有没有不下雨的日子，龙在最近的一天喝光湖里的水。刚开始本来想一边读取一边处理的，但是函数跳转的时候流操作对一个数读取了两次，导致直接出错了，最后就先把输入存下来了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <string>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>
#include <bitset>
#include <cassert>
#include <cmath>
#include <functional>

using namespace std;

const int maxn = 1000005;
int n, m, A[maxn], ans[maxn], full[maxn];

void init()
{
	memset(ans, 0, sizeof(ans));
	memset(full, 0, sizeof(ans));
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		cin >> A[i];
	}
}

void solve()
{
	set<int> E;
	bool err = false;
	for (int i = 1; i <= m && !err; i++) {
		if (A[i] > 0) {
			// 查找当前湖最后灌满的日子之后有没有不下雨的日子
			set<int>::iterator it = E.lower_bound(full[A[i]]);
			if (it != E.end() && *it > full[A[i]]) {
				ans[*it] = A[i];
				E.erase(it);
				full[A[i]] = i; // 湖满水的日子
			}
			else {
				err = true;
			}
		}
		else {
			E.insert(i);
		}
	}
	if (err) {
		cout << "NO\n";
	}
	else {
		cout << "YES\n";
		bool flag = false;
		for (int i = 1; i <= m; i++) {
			if (A[i] == 0) {
				if (flag) {
					cout << ' ';
				}
				flag = true;
				cout << ans[i];
			}
		}
		cout << endl;
	}
}

int main()
{
	ios::sync_with_stdio(false);
	int T;
	cin >> T;
	while (T--) {
		init();
		solve();
	}

	return 0;
}