HDU's ACM 1009 FatMouse' Trade

来源：互联网发布：mysql text使用编辑：程序博客网时间：2024/04/30 19:37

原题链接：HDU's ACM 1009 FatMouse' Trade

分析：用贪心算法解决。读入数组之后，按照J/F大小降序排列，正确性如下：a.若A'是问题P(A,W)的一个最优解，则a0总可以包含在A'中：i)若a0已在A'中，则得证；ii)若a0不在A'中，则取其中任意一个ak替换为a0后，A''不小于A';b.设A'是P(A,W)的一个最优解，且a0属于A'，则A'\{a0}也是P({A\{a0},W)的一个最优解：假设命题不成立，则存在A"\{a0}是最优解，则A"才是P(A,W)的最优解，与假设矛盾。综上，反复运用结论a,b，可知{a0...ak}为P(A,W)的最优解

AC Code:

#include <stdio.h>#include <string.h>#define MAXN 1000 + 10typedef struct Node{int J;int F;double weight;} Node;Node arr[MAXN];int Partition(Node arr[], int left, int right){int i, j;Node tmp;if(left < right){i = left;j = right;tmp.J = arr[i].J;tmp.F = arr[i].F;tmp.weight = arr[i].weight;while(i<j) {while(i<j && tmp.weight>arr[j].weight)j--;if(i<j) {arr[i].J = arr[j].J;arr[i].F = arr[j].F;arr[i].weight = arr[j].weight;i++;}while(i<j && tmp.weight<=arr[i].weight)++i;if(i<j){arr[j].J = arr[i].J;arr[j].F = arr[i].F;arr[j].weight = arr[i].weight;j--;}arr[i].J = tmp.J;arr[i].F = tmp.F;arr[i].weight = tmp.weight;}return i;}return -1;}void QuickSort(Node arr[], int left, int right){int pos;if(left < right){pos = Partition(arr, left, right);QuickSort(arr, left, pos-1);QuickSort(arr, pos+1, right);}}int main(){int M, N;int i;double res;#ifdef DEBUGfreopen("1009.in", "r", stdin);#endifwhile(scanf("%d%d", &M, &N) == 2){memset(arr, 0, MAXN*sizeof(Node));if(M == -1 && N == -1)break;for(i=1;i<=N;++i){scanf("%d%d", &arr[i].J, &arr[i].F);arr[i].weight = 1.0*arr[i].J/arr[i].F;}QuickSort(arr, 1, N);res = 0;i = 1;while(M && N-i+1){if(M>=arr[i].F){res += arr[i].J;M -= arr[i].F;} else {res += arr[i].J * 1.0 * M / arr[i].F;M = 0;}++i;}printf("%.3lf\n", res);}return 0;}

0 0