poj 3053 Fence Repair

题目链接http://poj.org/problem?id=3253

思路

题目与哈夫曼编码原理相同,使用优先队列与贪心思想;读入数据在优先队列中,弹出两个数计算它们的和,再压入队列中;

代码:

#include <iostream>
#include <queue>
using namespace std;

struct cmp
{
    bool operator() (long long a, long long b)
    {
        return a > b;
    }
};

int main()
{
    priority_queue<long long, vector<long long>, cmp> Q;
    long long n, ans = 0;

    cin >> n;
    for (int i = 0; i < n; ++i)
    {
        long long num;

        cin >> num;
        Q.push(num);
    }

    long long a, b;
    while (Q.size() != 1)
    {
        a = Q.top(); Q.pop();
        b = Q.top(); Q.pop();
        ans += a + b;
        Q.push(a + b);
    }

    cout << ans << endl;

    return 0;
}
时间: 2024-12-13 03:20:02

poj 3053 Fence Repair的相关文章

POJ 3253 Fence Repair (优先队列)

POJ 3253 Fence Repair (优先队列) Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needsN (1 ≤ N ≤ 20,000) planks of wood, each having some integer lengthLi (1 ≤ Li ≤ 50,000) units. He the

POJ 3253 Fence Repair 类似哈夫曼树的贪心思想

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24550   Accepted: 7878 Description Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000)

【优先队列/huffman】sdut 2848/poj 3253——Fence Repair

来源:点击打开链接 很久很久之前做过这个题,印象中是用优先队列来做,结果一写各种wa了..........翻之前的代码库,发现优先队列的定义出现了问题.. 因为数据很大需要每次都选取两个最短的进行拼装,所以用了优先队列,每两个小的构成父节点,然后把父节点放进去再找两个小的接起来.huffmanTree的逆向思维,接到最后那一个就是最后的答案了. #include <iostream> #include <queue> #include <vector> #include

poj 3253 Fence Repair(优先队列+哈夫曼树)

题目地址:POJ 3253 哈夫曼树的结构就是一个二叉树,每一个父节点都是两个子节点的和.这个题就是可以从子节点向根节点推. 每次选择两个最小的进行合并.将合并后的值继续加进优先队列中.直至还剩下一个元素为止. 代码如下: #include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #include <cty

POJ 3253 Fence Repair(哈夫曼树)

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 26167   Accepted: 8459 Description Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000)

哈夫曼树 POJ 3253 Fence Repair

竟然做过原题,一眼看上去竟然没感觉... 哈夫曼树定义:给定n个权值作为n个叶子结点,构造一棵二叉树,若带权路径长度达到最小,称这样的二叉树为最优二叉树,也称为哈夫曼树(Huffman tree).哈夫曼树是带权路径长度最短的树,权值较大的结点离根较近. 1.路径和路径长度 在一棵树中,从一个结点往下可以达到的孩子或孙子结点之间的通路,称为路径.通路中分支的数目称为路径长度.若规定根结点的层数为1,则从根结点到第L层结点的路径长度为L-1. 2.结点的权及带权路径长度 若将树中结点赋给一个有着某

POJ 3253 Fence Repair(优先队列,哈夫曼树)

题目 //做哈夫曼树时,可以用优先队列(误?) //这道题教我们优先队列的一个用法:取前n个数(最大的或者最小的) //哈夫曼树 //64位 //超时->优先队列,,,, //这道题的优先队列用于取前2个小的元素 #include <iostream> #include<stdio.h> #include<string.h> #include<algorithm> #include<queue> using namespace std; _

[ACM] POJ 3253 Fence Repair (Huffman树思想,优先队列)

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 25274   Accepted: 8131 Description Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000)

poj 3253 Fence Repair

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 40465   Accepted: 13229 Description Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000)