习题6-7(uva-804)

最新推荐文章于 2023-11-11 17:29:39 发布

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本文介绍了一个基于规则驱动的令牌更新系统实现方法，通过定义输入输出规则来更新令牌值，实现了状态转换，并能判断系统在指定界限内是否仍然存活。

#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 = 105;

int token[maxn],n,m,bound;
struct Rules {
	map<int, int> in,out;
}rules[maxn];

bool RunOnce() {
	
	for (int i = 1; i <= m; ++i) {
		bool isok = true;
		for (auto it = rules[i].in.begin(); it != rules[i].in.end(); ++it) {
			if (token[it->first] < it->second) {
				isok = false;
				break;
			}
		}
		if (isok) {
			for (auto it = rules[i].in.begin(); it != rules[i].in.end(); ++it) {
				token[it->first] -= it->second;
			}
			for (auto it = rules[i].out.begin(); it != rules[i].out.end(); ++it) {
				token[it->first] += it->second;
			}
			return true;
		}
	}
	return false;
}
int main()
{
	int t,cnt,id = 0;
	while (cin >> n && n) {

		for (int i = 1; i <= n; ++i)
			cin >> token[i];

		cin >> m;

		for (int i = 1; i <= m; ++i) {
			rules[i].in.clear();
			rules[i].out.clear();

			while (cin >> t && t) {
				if (t < 0) {
					rules[i].in[-t]++;
				}
				else {
					rules[i].out[t]++;
				}
			}
		}
		cin >> bound;
		
		for (cnt = 1; cnt <= bound; ++cnt) {
			if(!RunOnce()) break;
		}

		if (cnt <= bound) cout << "Case " << ++id << ": dead after " << cnt-1 << " transitions" << endl;
		else cout << "Case " << ++id << ": still live after " << bound << " transitions" << endl;

		cout << "Places with tokens:";

		for (int i = 1; i <= n; ++i) {
			if (token[i]) {
				cout << " " << i << " (" << token[i] << ")";
			}
		}
		cout << endl << endl;
	}
	return 0;
}