POJ 3253 Fence Repair

题目链接：
http://poj.org/problem?id=3253
解题思路：

朴素方法代码又长，而且又不易懂。。。其实只需从板的集合里取出最短的两块，并且把长度为两块板长度之和的板加入集合中即可，因此如果使用优先队列就可以高效地实现。时间复杂复为O（N log N）.

AC代码（朴素方法）：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long ll;
int n,a[20005];

void solve(){
    ll ans = 0;
    //直到计算到木板为1快时为止
    while(n > 1){
        //求出最短的板mii1和次短的板mii2
        int mii1 = 0,mii2 = 1;
        if(a[mii1] > a[mii2])
            swap(mii1,mii2);
        for(int i = 2; i < n; i++){
            if(a[i] < a[mii1]){
                mii2 = mii1;
                mii1 = i;
            }
            else if(a[i] < a[mii2]){
                mii2 = i;
            }
        }
        //将两块板拼合
        int t = a[mii1] + a[mii2];
        ans += t;

        if(mii1 == n-1)
            swap(mii1,mii2);
        a[mii1] = t;
        a[mii2] = a[n-1];
        n--;
    }
    printf("%lld\n",ans);
}

int main(){
    while(~scanf("%d",&n)){
        for(int  i = 0; i < n; i++)
            scanf("%d",&a[i]);
        solve();
    }
    return 0;
}

AC代码（优先队列）：

<pre name="code" class="cpp">#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;

typedef long long ll;
int n,x;
priority_queue<int,vector<int>,greater<int> > q;

void solve(){
    ll ans = 0;
    while(q.size() > 1){
        int l1 = q.top();
        q.pop();
        int l2 = q.top();
        q.pop();
        //printf("%d %d\n",l1,l2);
        ans += l1+l2;
        q.push(l1+l2);
    }
    printf("%lld\n",ans);
}

int main(){
    while(~scanf("%d",&n)){
        while(!q.empty())
            q.pop();
        for(int  i = 0; i < n; i++){
            scanf("%d",&x);
            q.push(x);
        }
        solve();
    }
    return 0;
}