PKU ACM poj 2231 源代码

简单题，关键是提高效率。

使用暴力计算，耗时1000ms，改进算法后为63ms，然后再改进程序中的细节，进而优化到16ms。

 

 

 

 

源代码（16MS）：

/*Moo VolumeTime Limit: 1000MS Memory Limit: 65536K Total Submissions: 12448 Accepted: 3582 DescriptionFarmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise. FJ's N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty animals. Every pair of cows simultaneously carries on a conversation (so every cow is simultaneously MOOing at all of the N-1 other cows). When cow i MOOs at cow j, the volume of this MOO must be equal to the distance between i and j, in order for j to be able to hear the MOO at all. Please help FJ compute the total volume of sound being generated by all N*(N-1) simultaneous MOOing sessions.Input* Line 1: N * Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).OutputThere are five cows at locations 1, 5, 3, 2, and 4.Sample Input515324Sample Output40HintINPUT DETAILS: There are five cows at locations 1, 5, 3, 2, and 4. OUTPUT DETAILS: Cow at 1 contributes 1+2+3+4=10, cow at 5 contributes 4+3+2+1=10, cow at 3 contributes 2+1+1+2=6, cow at 2 contributes 1+1+2+3=7, and cow at 4 contributes 3+2+1+1=7. The total volume is (10+10+6+7+7) = 40.*/#include "stdlib.h"#include "stdio.h"#define MAX_COW_NUM 10000typedef struct _COW_ST_ {int iLocation;/*in the range 0..1,000,000,000*/}COW_ST;typedef struct _MOO_VOLUME_ST_{__int64 i64TotalVolume;int iCowNum;COW_ST astCow[MAX_COW_NUM];}MOO_VOLUME_ST;MOO_VOLUME_ST gstMooVolume;void CowInput(void){int iLoopCow;scanf("%d",&gstMooVolume.iCowNum); for(iLoopCow = 0; iLoopCow < gstMooVolume.iCowNum; iLoopCow++){scanf("%d",&gstMooVolume.astCow[iLoopCow].iLocation);}return;}int CowCmp(const void *a,const void *b){if (((COW_ST *)a)->iLocation > ((COW_ST *)b)->iLocation){return 1;}else{return -1;}}void CalculateMooVolume(void){int iLoopCow;__int64 i64LastCowVolume = 0;/*get 1st cow's moo volume*/for(iLoopCow = 1; iLoopCow < gstMooVolume.iCowNum; iLoopCow++){i64LastCowVolume += (gstMooVolume.astCow[iLoopCow].iLocation - gstMooVolume.astCow[0].iLocation);}gstMooVolume.i64TotalVolume = i64LastCowVolume;/*get 2nd~Nth cow's moo volume*/for(iLoopCow = 1; iLoopCow < gstMooVolume.iCowNum ; iLoopCow++){i64LastCowVolume += ( iLoopCow+iLoopCow - gstMooVolume.iCowNum)*((__int64)(gstMooVolume.astCow[iLoopCow].iLocation - gstMooVolume.astCow[iLoopCow-1].iLocation));gstMooVolume.i64TotalVolume += i64LastCowVolume;}return;}void CalculateMooVolume_Common(void){int iLoopCow;int iLoopCow_1;gstMooVolume.i64TotalVolume = 0;for(iLoopCow = 0; iLoopCow < gstMooVolume.iCowNum; iLoopCow++){for(iLoopCow_1 = 0; iLoopCow_1 < gstMooVolume.iCowNum; iLoopCow_1++){gstMooVolume.i64TotalVolume += abs(gstMooVolume.astCow[iLoopCow].iLocation - gstMooVolume.astCow[iLoopCow_1].iLocation);}}return;}int MooVolumeMain(void){CowInput();qsort(gstMooVolume.astCow ,gstMooVolume.iCowNum ,sizeof(COW_ST ),CowCmp);CalculateMooVolume();//CalculateMooVolume_Common();printf("%I64d/n", gstMooVolume.i64TotalVolume );return 0;}