51nod 1117 聪明的木匠 (贪心）

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1117

跟挑战程序书上例题一样，将要切割的n断木板，分别对应二叉树树的叶子节点，则切割的总开销为木板的长度×节点的深度，可以画图理解，那么最短的木板（L1）应当是深度最大的叶子节点之一，次短的板（L2）和它是兄弟节点，由一块长度是（L1+L2） 的木板切割而来，这样可以变成（ｎ－１）块木板，然后递归求解到ｎ＝＝１。

书上贪心部分用的是O（ｎ×ｎ）　的算法　　在这里会超时，需要２.４节优先队列O（NｌｏｇN）的算法。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>

 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1000000000
 #define N 50005
 using namespace std;

 int L[N];
 int main()
 {
    // Read();
     int n;
     long long ans=;
     scanf("%d",&n);
     for(int i=;i<n;i++) scanf("%d",&L[i]);
     while(n>)
     {
         int m1=,m2=;
         if(L[m1]>L[m2]) swap(m1,m2);
         for(int i=;i<n;i++)
         {
             if(L[m1]>L[i])
             {
                 m2=m1;
                 m1=i;
             }
             else if(L[m2]>L[i])
             {
                 m2=i;
             }
         }
         //printf("%d %d\n",L[m1],L[m2]);
         int t=L[m1]+L[m2];
         ans+=t;
         if(m1==n-) swap(m1,m2);
         L[m1]=t;
         L[m2]=L[n-];
         n--;
     }
     printf("%lld\n",ans);
     return ;
 }

优化后的代码：每次只需要选择长度最小的两块木板，则用优先队列从小到大排序，每次把取出来的两块木板之和压进队列即可。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 //#include <map>
 #include <queue>

 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1000000000
 #define N 50005
 using namespace std;

 int L[N];
 priority_queue<int, vector<int>, greater<int> >que;
 int main()
 {
     //Read();
     int n;
     scanf("%d",&n);
     for(int i=;i<n;i++)
     {
         scanf("%d",&L[i]);
         que.push(L[i]);
     }
     ll ans=;
     while(que.size()>)
     {
         int l1,l2;
         l1=que.top();
         que.pop();
         l2=que.top();
         que.pop();
         ans+=l1+l2;
         que.push(l1+l2);
     }
     printf("%lld\n",ans);
     return ;
 }