zoukankan      html  css  js  c++  java
  • UVALive 7274 Canvas Painting (优先队列)

    Canvas Painting

    题目链接:

    http://acm.hust.edu.cn/vjudge/contest/127406#problem/C

    Description

    http://7xjob4.com1.z0.glb.clouddn.com/a4717ad58f73aa6ff84a1ab3f051c3f8

    Input

    The first line consists of a single integer T, the number of test cases. Each test case is composed by two lines. The first line consists of a single integer N representing the number of canvasses. The next line contains N space separated integers representing the sizes of the canvasses. Constraints: 1 ≤ T ≤ 100 Number of test cases. 1 ≤ Ni ≤ 100 000 Number of canvasses in the i th test case. 1 ≤ s ≤ 100 000 Size of each canvas. 1 ≤ ∑Ti=1 Ni ≤ 100 000 Number of canvasses over all test cases in one test file.

    Output

    The output contains T lines, one for each test case: the minimum amount of ink the machine needs in order to have all canvasses with different colors.

    Sample Input

    2 3 7 4 7 4 5 3 7 5

    Sample Output

    29 40
    ##题意: 给出N张白布(顺序不定). 每次选出其中同一种颜色的若干张布染上某种跟之前不同的色,这种颜色剩下的布染上另一种颜色. 每次染色的花费是布的大小. 求要将N张布都染成不同的颜色的最小花费.
    ##题解: 一开始想的是面积大的布染尽量少的次数,先降序排列,对后缀和求和. 这个思路并不正确. (比如样例2) 这个问题反过来看就比较简单了: 最后的结果是N张颜色各异的布,反向过程是每次选出两种颜色不同的布刷成同一颜色. 这样一来,每次操作都使得集合的大小减一. 所以总次数固定是N-1. 对于每一次操作,选择最小的两张布染色一定是最小花费. 而每次的最小花费和就是总的最小花费.
    维护一个优先队列,把所有大小都加进去并升序排列. 每次弹出最小的两个元素,计数并把和再push进去参与比较.

    ##代码: ``` cpp #include #include #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 101000 #define mod 100000007 #define inf 0x3f3f3f3f #define mid(a,b) ((a+b)>>1) #define IN freopen("in.txt","r",stdin); using namespace std;

    priority_queue<LL, vector, greater > pq;

    int main(int argc, char const *argv[])
    {
    //IN;

    int t; cin >> t;
    while(t--)
    {
        int n; scanf("%d", &n);
    
        while(!pq.empty()) pq.pop();
        for(int i=1; i<=n; i++) {
            LL x; scanf("%lld", &x);
            pq.push(x);
        }
    
        LL ans = 0;
        while(pq.size() >= 2) {
            LL cur = pq.top(); pq.pop();
            cur += pq.top(); pq.pop();
            ans += cur;
            pq.push(cur);
        }
    
        printf("%lld
    ", ans);
    }
    
    return 0;
    

    }

  • 相关阅读:
    cocos2d-x 坐标系
    Linux 用户和用户组
    Linux 挂载分区 + swap 分区
    Linux 分区 磁盘分区与格式化
    Linux MBR分区(重点知识)
    Linux -磁盘管理 ip http://blog.csdn.net/xh16319/article/details/17272455
    Linux 底行模式常用命令
    Linux Bash 通配符
    Linux Bash 的基本功能 管道符
    Linux Bash的基本功能 输出重定向
  • 原文地址:https://www.cnblogs.com/Sunshine-tcf/p/5769161.html
Copyright © 2011-2022 走看看