Fence Repair (二叉树求解)(优先队列,先取出小的)
题目链接:http://poj.org/problem?id=3253
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 61005 | Accepted: 20119 |
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) planks of wood, each having some integer length Li (1 ≤ Li ≤ 50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths Li). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made; you should ignore it, too.
FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to Farmer Don's Farm with this long board and politely asks if he may borrow a saw.
Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.
Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will result in different charges since the resulting intermediate planks are of different lengths.
Input
Lines 2..N+1: Each line contains a single integer describing the length of a needed plank
Output
Sample Input
3
8
5
8
Sample Output
34
Hint
The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).
#include<iostream>
#include<string.h>
#include<map>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
typedef long long ll;
using namespace std;
const ll mod=1e9+;
const int maxn=2e4+;
const int maxk=1e4+;
const int maxx=1e4+;
const ll maxe=+;
#define INF 0x3f3f3f3f3f3f
#define Lson l,mid,rt<<1
#define Rson mid+1,r,rt<<1|1
int n;
int a[maxn];
void solve()
{
ll ans=;
while(n>)
{
int mi1=,mi2=;
if(a[mi1]>a[mi2]) swap(mi1,mi2);
for(int i=;i<n;i++)
{
if(a[i]<a[mi1])
{
mi2=mi1;
mi1=i;
}
else if(a[i]<a[mi2])
{
mi2=i;
}
}
int l=a[mi1]+a[mi2];
ans+=l;
if(a[mi1]==n-) swap(mi1,mi2);
a[mi1]=l;
a[mi2]=a[n-];
n--;
}
cout<<ans<<endl;
}
int main()
{
cin>>n;
for(int i=;i<n;i++)
{
cin>>a[i];
}
sort(a,a+n);
solve();
return ;
}
上面的算法复杂度是n*n,其实还有一种更快的方法,原理跟上面的一样,但是下面的算法用到优先队列,所以把复杂度降到nlogn
看代码
#include<iostream>
#include<string.h>
#include<map>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
typedef long long ll;
using namespace std;
const ll mod=1e9+;
const int maxn=2e4+;
const int maxk=1e4+;
const int maxx=1e4+;
const ll maxe=+;
#define INF 0x3f3f3f3f3f3f
#define Lson l,mid,rt<<1
#define Rson mid+1,r,rt<<1|1
int n;
int a[maxn];
priority_queue<int,vector<int>,greater<int> >que;//声明一个从小到大取出数值的优先队列
void solve()
{
ll ans=;
for(int i=;i<n;i++)
que.push(a[i]);//全部存入队列
while(n>)
{
int l1,l2;
l1=que.top();
que.pop();
l2=que.top();
que.pop();
int t=l1+l2;
ans+=t;
que.push(t);
n--;
}
cout<<ans<<endl;
}
int main()
{
cin>>n;
for(int i=;i<n;i++)
{
cin>>a[i];
}
// sort(a,a+n);
solve();
return ;
}
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