该题与石子合并的区别:石子合并是两两相邻才能合并,这题是任意两点合并
该题的思路:每次合并最小的两个点
- #include
- #include
- #include
-
- using namespace std;
-
- int main()
- {
- int n;
- scanf("%d", &n);
-
- priority_queue<int, vector<int>, greater<int>> heap;//小根堆,所以堆头是最小值
- while (n -- )
- {
- int x;
- scanf("%d", &x);
- heap.push(x);
- }
-
- int res = 0;//答案
- while (heap.size() > 1)//只要超过两个数,就两个合并。因为是小根堆,每次合并的都是最小的两个
- {
- int a = heap.top(); heap.pop();
- int b = heap.top(); heap.pop();
- res += a + b;
- heap.push(a + b);
- }
-
- printf("%d\n", res);
- return 0;
- }