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POJ 3253 Fence Repair 哈夫曼树/优先级队列

2013年08月16日 ⁄ 综合 ⁄ 共 1487字 ⁄ 字号 评论关闭

题意:将一根长为l的木棍锯成n段,使l1+l2+l3+....= l,  且每次锯木棍所需花费等于木棍长度。求最小花费。
题解:学习优先级队列。

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

struct cmp
{
	bool operator () ( const int &a, const int &b )
	{
		return a > b ;
	}
};

int main()
{
	
	int n;
	__int64 ans = 0, first, second, temp;
	priority_queue<int,vector<int>,cmp> Q;

	scanf("%d",&n);
	for ( int i = 0; i < n; ++i )
	{
		scanf("%d",&temp);
		Q.push ( temp );
	}

	while ( Q.size () != 1 )
	{
		first = Q.top (); Q.pop ();
		second = Q.top(); Q.pop ();
		temp = first + second;
		Q.push ( temp );
		ans += temp;
	}
	printf("%I64d\n",ans);
	return 0;
}

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

struct node
{
	__int64 value;
	bool operator < ( const node &a ) const
	{
		return value > a.value;
	}
} temp;  // 将temp定义在main函数里为什么就是错的呢?

int main()
{	
	int n;
	node first, second;
	priority_queue<node> Q;

	scanf("%d",&n);
	for ( int i = 0; i < n; ++i )
	{
		scanf("%d",&temp.value);
		Q.push ( temp );
		printf("%I64d ", Q.top().value);
	}

	__int64 ans = 0;
	while ( Q.size () != 1 )
	{
		first = Q.top(); Q.pop ();
		second = Q.top(); Q.pop ();
		temp.value = first.value + second.value;
		Q.push ( temp );
		ans += temp.value;
	}
	printf("%I64d\n",ans);
	return 0;
}

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

int array[20005], n;

int cmp ( const void *a, const void *b )
{
	return *((int*)a) - *((int*)b);
}

void in ( int num, int start )
{
	int i = start;
	while ( i < n && array[i] < x )
	{
		array[i-1] = array[i];
		++i;
	}
	array[i-1] = x;
}

int main()
{
	int temp, i;
	scanf("%d",&n);
	for ( i = 0; i < n; ++i )
		scanf("%d", array+i);
	qsort(array,n,sizeof(int),cmp);
	__int64 ans = 0; 
	i = 0;
	while ( i < n-1 ) //当数组只剩下一个元素时则跳出。(因为该值在上一次循环时已经用到)
	{
		temp = array[i] + array[i+1];
		in ( temp, i+2 )  //每次扔掉数组的前两个元素,并将temp的值按排序规律插到特定位置
		ans += temp;
		++i;
	}
	printf("%I64d\n",ans);
	return 0;
}

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