zoukankan      html  css  js  c++  java
  • Day3-E-New Year Snowmen CodeForces140C

    As meticulous Gerald sets the table and caring Alexander sends the postcards, Sergey makes snowmen. Each showman should consist of three snowballs: a big one, a medium one and a small one. Sergey's twins help him: they've already made n snowballs with radii equal to r1r2, ..., rn. To make a snowman, one needs any three snowballs whose radii are pairwise different. For example, the balls with radii 1, 2 and 3 can be used to make a snowman but 2, 2, 3 or 2, 2, 2 cannot. Help Sergey and his twins to determine what maximum number of snowmen they can make from those snowballs.

    Input

    The first line contains integer n (1 ≤ n ≤ 105) — the number of snowballs. The next line contains n integers — the balls' radii r1r2, ..., rn (1 ≤ ri ≤ 109). The balls' radii can coincide.

    Output

    Print on the first line a single number k — the maximum number of the snowmen. Next k lines should contain the snowmen's descriptions. The description of each snowman should consist of three space-separated numbers — the big ball's radius, the medium ball's radius and the small ball's radius. It is allowed to print the snowmen in any order. If there are several solutions, print any of them.

    Examples

    Input
    7
    1 2 3 4 5 6 7
    Output
    2
    3 2 1
    6 5 4
    Input
    3
    2 2 3
    Output
    0

    思路:贪心,每次选出还剩的最大的三个组成一个雪人,用优先队列为数据结构,代码如下:
    struct Node {
        int id, sum;
        Node(int _id = 0, int _sum = 0) : id(_id), sum(_sum){}
        bool operator<(const Node &a) const {
            return sum < a.sum;
        }
    };
    
    struct Snowman {
        int a, b, c;
        Snowman(int _a, int _b, int _c) : a(_a), b(_b), c(_c){}
    };
    
    bool comp(int a,int b) {
        return a > b;
    }
    map<int, int> Cache;
    vector<Snowman> res;
    int n;
    
    int main() {
        scanf("%d", &n);
        priority_queue<Node> q;
        for (int i = 0; i < n; ++i) {
            int tmp;
            scanf("%d", &tmp);
            Cache[tmp]++;
        }
        for(auto i = Cache.begin(); i != Cache.end(); ++i)
            q.push(Node(i->first,i->second));
        while(q.size() > 2) {
            Node t1, t2, t3;
            t1 = q.top(), q.pop();
            t2 = q.top(), q.pop();
            t3 = q.top(), q.pop();
            res.push_back(Snowman(t1.id, t2.id, t3.id));
            t1.sum--, t2.sum--, t3.sum--;
            if(t1.sum > 0) q.push(t1);
            if(t2.sum > 0) q.push(t2);
            if(t3.sum > 0) q.push(t3);
        }
        printf("%d
    ", res.size());
        for (int i = 0; i < res.size(); ++i) {
            int tmp[3];
            tmp[0] = res[i].a, tmp[1] = res[i].b, tmp[2] = res[i].c;
            sort(tmp, tmp + 3, comp);
            for(int j = 0; j < 3; ++j) {
                if(j)
                    printf(" ");
                printf("%d", tmp[j]);
            }
            printf("
    ");
        }
        return 0;
    }
    View Code
     
  • 相关阅读:
    ubuntu 用shell脚本实现将当前文件夹下全部文件夹中的某一类文件复制到同一文件夹下
    读书笔记-2java虚拟机的可达性算法与finalize方法
    find the longest of the shortest (hdu 1595 SPFA+枚举)
    杭电 2176 取(m堆)石子游戏(博弈)
    MVC框架的优缺点
    Wireshark-TCP协议分析(包结构以及连接的建立和释放)
    Ubuntu安装教程--Win7系统中含100M保留分区
    eclipse新建android项目出现非常多错误
    关于简单的加密和解密算法
    在一台server上部署多个Tomcat
  • 原文地址:https://www.cnblogs.com/GRedComeT/p/11239776.html
Copyright © 2011-2022 走看看