zoukankan      html  css  js  c++  java
  • 【习题 6-10 UVA

    【链接】 我是链接,点我呀:)
    【题意】

    在这里输入题意

    【题解】

    发牌的那个牌堆用一个deque,7个牌堆用vector来模拟。 然后按照题意模拟就好。 不难。

    【代码】

    /*
    1.Shoud it use long long ?
    2.Have you ever test several sample(at least therr) yourself?
    3.Can you promise that the solution is right? At least,the main ideal
    4.use the puts("") or putchar() or printf and such things?
    5.init the used array or any value?
    6.use error MAX_VALUE?
    7.use scanf instead of cin/cout?
    8.whatch out the detail input require
    */
    #include <bits/stdc++.h>
    using namespace std;
    
    const int N = 8;
    
    deque <int> dl;
    map <vector <int>, int > dic;
    vector <int> pile[N];
    bool bo[N];
    int cur;
    
    int judge() {
    	vector <int> v;
    	v.clear();
    	for (auto i = dl.begin(); i != dl.end(); i++) {
    		v.push_back((*i));
    	}
    
    	for (int i = 1; i <= 7; i++) {
    		v.push_back(0);
    		for (int x : pile[i]) {
    			v.push_back(x);
    		}
    	}
    
    	if (dic.find(v) != dic.end()) return 1;
    	dic[v] = 1;
    	return 0;
    }
    
    bool can() {
    	int len = pile[cur].size();
    	if (len<3) return false;
    
    	int ju = pile[cur][0] + pile[cur][1] + pile[cur][len-1];
    
    	if (ju == 10 || ju == 20 || ju == 30) {
    		dl.push_back(pile[cur][0]);
    		dl.push_back(pile[cur][1]);
    		dl.push_back(pile[cur][len-1]);
    
    		pile[cur].erase(pile[cur].begin());
    		pile[cur].erase(pile[cur].begin());
    		pile[cur].erase(pile[cur].end() - 1);
    		return true;
    	}
    
    	ju = pile[cur][0] + pile[cur].back() + pile[cur][len - 2];
    	if (ju == 10 || ju == 20 || ju == 30) {
    		dl.push_back(pile[cur][0]);
    		dl.push_back(pile[cur][len - 2]);
    		dl.push_back(pile[cur][len - 1]);
    
    		pile[cur].erase(pile[cur].begin());
    		pile[cur].erase(pile[cur].end() - 1);
    		pile[cur].erase(pile[cur].end() - 1);
    		return true;
    	}
    
    	ju = pile[cur][len - 3] + pile[cur][len - 2] + pile[cur].back();
    	if (ju == 10 || ju == 20 || ju == 30) {
    		dl.push_back(pile[cur][len - 3]);
    		dl.push_back(pile[cur][len - 2]);
    		dl.push_back(pile[cur][len - 1]);
    
    		pile[cur].erase(pile[cur].end() - 1);
    		pile[cur].erase(pile[cur].end() - 1);
    		pile[cur].erase(pile[cur].end() - 1);
    		return true;
    	}
    
    	return false;
    }
    
    int main() {
    	#ifdef LOCAL_DEFINE
    		freopen("F:\c++source\rush_in.txt", "r", stdin);
    				freopen("F:\c++source\rush_out.txt", "w", stdout);
    	#endif
    	ios::sync_with_stdio(0), cin.tie(0);
    	int x;
    	while (cin >> x && x) {
    		for (int i = 1; i <= 7; i++) bo[i] = true;
    		for (int i = 1; i <= 7; i++) pile[i].clear();
    		dic.clear();
    		while (!dl.empty()) dl.pop_back();
    
    		dl.push_back(x);
    		for (int i = 1; i <= 51; i++) {
    			cin >> x;
    			dl.push_back(x);
    		}
    
    		judge();
    		int tot = 0;
    
    		int ans;
    		cur = 1;
    		while (1) {
    			if (dl.empty()) { //lose
    				ans = -1;
    				break;
    			}
    			int cntalive = 0;
    			for (int i = 1; i <= 7; i++) cntalive += bo[i];
    			if (cntalive == 0) {//win
    				ans = 1;
    				break;
    			}
    
    			//deal one card
    			if (cur > 7) cur = 1;
    			while (!bo[cur]) {
    				cur++;
    				if (cur > 7) cur = 1;
    			}
    
    			pile[cur].push_back(dl.front());
    			dl.pop_front();
    
    			do {
    			} while (can());
    
    			if (pile[cur].empty()) bo[cur] = 0;
    
    			tot++;
    
    			int temp = judge();
    			if (temp == 1) {
    				ans = 0;//loop;
    				break;
    			}
    			cur++;
    		}
    
    		if (ans == 0) {
    			cout << "Draw: ";
    		}
    		else if (ans == 1) {
    			cout << "Win : ";
    		}
    		else if (ans == -1) {
    			cout << "Loss: ";
    		}
    		cout << tot << endl;
    	}
    	return 0;
    }
    
  • 相关阅读:
    django中使用redis保存session(转)
    Python操作Redis(转)
    Django 中 app_name (应用命名空间) 和 namespace (实例命名空间) 的区别
    零开始Android逆向教程(一)——初探Android逆向
    python基础网络编程--转
    安全建设之平台搭建
    从信息安全弃坑到虚拟化的悲哀和无奈之感
    Apache 性能配置优化
    反爬虫和抗DDOS攻击技术实践
    DockerCon2017前瞻
  • 原文地址:https://www.cnblogs.com/AWCXV/p/7907821.html
Copyright © 2011-2022 走看看