poj 2442 堆的应用（m组数取前n小）

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题意：有m组数，每组n个。从每组中取一个数求和，可以得到n^m个和，求这n^m个数中的最小的n个

思路：一组一组数据的处理，建立大顶堆。heap为大顶堆，s数组保存每组数的读入，temp是处理到当前组的前n小的数的递增序列。

#include <stdio.h>#include <string.h>#include <stdlib.h>#define N 2002int heap[N],s[N],temp[N];int T,n,m;int cmp(const int *a,const int *b){return (*a)-(*b);}void adjust(int i){int temp = heap[i];i <<= 1;while(i<=n){if(i+1<=n && heap[i+1]>heap[i])i++;if(temp > heap[i])break;heap[i/2] = heap[i];i <<= 1;}heap[i/2] = temp;}int main(){freopen("a.txt","r",stdin);scanf("%d",&T);while(T--){int i,j,loop ;scanf("%d %d",&m,&n);loop = m-1;for(i = 1;i<=n;i++){scanf("%d",&s[i]);heap[i] = s[i];temp[i] = s[i];}for(i = n/2;i>=1;i--)//建立大顶堆adjust(i);qsort(temp+1,n,sizeof(int),cmp);while(loop--){for(i = 1;i<=n;i++)scanf("%d",&s[i]);qsort(s+1,n,sizeof(int),cmp);for(i = 1;i<=n;i++)heap[i] += s[1];for(i = 2;i<=n;i++){for(j = 1;j<=n;j++){if(s[i] + temp[j] < heap[1]){heap[1] = s[i]+temp[j];adjust(1);}elsebreak;}if(j==1)break;}for(i = 1;i<=n;i++)temp[i] = heap[i];qsort(temp+1,n,sizeof(int),cmp);}for(i = 1;i<=n;i++)printf("%d ",temp[i]);printf("\n");}return 0;}

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