POJ 3253 (哈弗曼树) ACM之路

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14281		Accepted: 4548

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
Line 1: One integer N, the number of planks
Lines 2.. N+1: Each line contains a single integer describing the length of a needed plank 【POJ 3253 (哈弗曼树)】Output
Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts
堆排序实现

#include #include#define clear(a,b) mesmet(a,b,sizeof(a)) #define maxn 20010long longn,hs = 0,heap[maxn]; void heapup(int p) { int q = p>>1,a = heap[p]; while(q) { if (a < heap[q]) heap[p] = heap[q]; else break; p = q; q = p >> 1; } heap[p] = a; }void addtoheap(int a) { heap[++hs] = a; heapup(hs); }void heapdown(int p) { int q = p<<1,a = heap[p]; while(q <= hs) { if (q <= hs && heap[q+1] < heap[q]) q++; if (heap[q] < a) heap[p] = heap[q]; else break; p = q; q = p << 1; } heap[p] = a; }int gettop() { int top = heap[1]; heap[1] = heap[hs--]; heapdown(1); return top; }int work() { long long m,i,tmp,t1,ans = 0; scanf("%lld",&m); for(i = 1; i <= m; i++) { scanf("%lld",&tmp); addtoheap(tmp); } for(i = 1; i < m; i++) { t1 = gettop(); t1 += gettop(); ans += t1; addtoheap(t1); } printf("%lld\n",ans); }int main() { work(); return 0; }