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 Volume Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12448 Accepted: 3582 Description Farmer 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). Output There are five cows at locations 1, 5, 3, 2, and 4. Sample Input 5 1 5 3 2 4 Sample Output 40 Hint INPUT 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 10000 typedef 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; }