(PAT 1125) Chain the Ropes (贪心+优先级队列的应用)

绝地灬酷狼 2022-03-22 13:14 206阅读 0赞

1125 Chain the Ropes （25 point(s)）

Given some segments of rope, you are supposed to chain them into one rope. Each time you may only fold two segments into loops and chain them into one piece, as shown by the figure. The resulting chain will be treated as another segment of rope and can be folded again. After each chaining, the lengths of the original two segments will be halved.

![rope.jpg][]

Your job is to make the longest possible rope out of N given segments.

### Input Specification: ###

Each input file contains one test case. For each case, the first line gives a positive integer N (2≤N≤104). Then N positive integer lengths of the segments are given in the next line, separated by spaces. All the integers are no more than 104.

### Output Specification: ###

For each case, print in a line the length of the longest possible rope that can be made by the given segments. The result must be rounded to the nearest integer that is no greater than the maximum length.

### Sample Input: ###

8
    10 15 12 3 4 13 1 15

### Sample Output: ###

解题思路:

用一个优先级队列进行维护,确保最小的元素在队首,每次从队列中挑出最小的两个元素,合并后放回队列

以此类推,最后得到的那个元素就是最大的

#include <iostream>
    #include <algorithm>
    #include <queue>
    using namespace std;
    const int MAXN = 10010;
    int ropes[MAXN];
    
    int main() {
    	int N;
    	cin >> N;
    	priority_queue<int,vector<int>,greater<int>> ropes_queue;
    	for (int i = 0; i < N; ++i) {
    		int num;
    		cin >> num;
    		ropes_queue.push(num);
    	}
    	while (!ropes_queue.empty()) {
    		if (ropes_queue.size() == 1) {
    			cout << ropes_queue.top() << endl;
    			system("PAUSE");
    			return 0;
    		}
    		int temA = ropes_queue.top();
    		ropes_queue.pop();
    		int temB = ropes_queue.top();
    		ropes_queue.pop();
    		double temC = (temA + temB)*1.0 / 2;  //四舍五入
    		ropes_queue.push((int)temC);
    	}
    	system("PAUSE");
    	return 0;
    }

[rope.jpg]: /images/20220322/6d6e3a8288c1461a96ecfeded70f23d3.png