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：

#include 
#include 
#include 
#include 
#include 
#include 
#include 

转载于:https://www.cnblogs.com/Sunshine-tcf/p/5769161.html