POJ 3253 Fence Repair (Mansha)

Topic Links:

http://poj.org/problem?id=3253

Main topic:

There is a stick, need to cut into n knots, each section has a fixed length, a stick of X-length into two segments, need to spend X, asked to cut into the required state to require the minimum cost?

Problem Solving Ideas:

Huffman number, the need for each section of the stick length of the tree as a node, the process of cutting the stick upside down, into the N-cut sticks connected, the two process costs are the same. According to the nature of Huffman, we know that the shortest two sticks together, put to the remaining n-2 root stick, and then select two the shortest connection up, and then put back, until all the wood roots are connected together OK.

Code:

1#include <cstdio>2#include <cstring>3#include <cstdlib>4#include <algorithm>5#include <iostream>6#include <cmath>7#include <queue>8 using namespacestd;9 Ten intMain () One { APriority_queue <int> Q;//priority queue, default big First out team, so we take the number of incoming team negative -     intN; -     Long Longsum, num, m; thescanf ("%d", &n); -      while(N--) -     { -scanf ("%d", &m); +Q.push (-M);//into the team -     } +sum =0; A      while(Q.size () >1)//the number of elements in the queue is greater than 1 at     { -num =0; -  -num + = Q.top ();//out of the team two shortest sticks - Q.pop (); -  innum + =q.top (); - Q.pop (); to  +sum + = num;//Link Up -Q.push (num);//into the team the     } *printf ("%lld\n", -sum); $     return 0;Panax Notoginseng}

POJ 3253 Fence Repair (Mansha)