PKU3399贪心

原题http://poj.org/problem?id=3399

Product

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2839		Accepted: 688		Special Judge

Description

There is an array of N integer numbers in the interval from -30000 to 30000. The task is to select K elements of this array with maximal possible product.

Input

The input consists of N + 1 lines. The first line contains N and K (1 ≤ K ≤ N ≤
100) separated by one or several spaces. The others contain values of array elements.

Output

The output contains a single line with values of selected elements separated by one space. These values must be in non-increasing order.

Sample Input

4 2
1
7
2
0

Sample Output

7 2

//第一次写的时候想分类讨论，发现这是一个很复杂的问题，需要考虑的问题有很多
//第二次的时候想到了用找的方法。但是比赛的时候感觉到有点麻烦。所以就没写。。。现在想想，挺后悔的。
//本题的思路是先按绝对值进行排序。然后进行查询。如果说碰到了负的先记录。如果碰到正的，就放入

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
using namespace std;
#define N 100 + 10

int abb(int a){
if(a  < 0)
return -a;
else
return a;
}

int cmp(int a,int b){
return abb(a) < abb(b);
}

int main(){
int n,k;
int i;
int num
,ans
,f[5];

while(~scanf("%d%d",&n,&k)){
memset(num,0,sizeof(num));
memset(ans,0,sizeof(ans));
memset(f,0,sizeof(f));
for(i=0;i<n;i++){
scanf("%d",&num[i]);
}
sort(num,num+n,cmp);
int l = 0;
int flag = 0;
for(i=n-1;i>=0&&l<k;i--){
if(num[i]<0 && flag==0){
f[0] = num[i];
flag = 1;
}
else if(num[i]<0 && flag==1 && l+2<=k){//如果放的下就直接放下
ans[l++] = f[0];
ans[l++] = num[i];
flag = 0;
}
else if(num[i]<0 && flag==1 && l+2>k){//如果不够了，那就就要进行判断。选择最优解
int j,x;
for(j=i-1;j>=0;j--){
if(num[j] >= 0){
break;
}
}
for(x=l-1;x>=0;x--){
if(ans[x] >= 0){//楼了个等于。。。。
break;
}
}
if(x>=0 && j>=0){
if(num[j]*ans[x] >= f[0]*num[i]){
ans[l++] = num[j];
}
else{
ans[l++] = num[i];
ans[x] = f[0];
}
}
flag = 0;
}
else if(num[i] > 0){
ans[l++] = num[i];
}
}
if(l < k){//如果案例为5 5 3 2 1 0 0
for(i=0;i<k;i++){
ans[i] = num[i];
}
}
sort(ans,ans+k);
for(i=k-1;i>=0;i--){
if(i != 0){
printf("%d ",ans[i]);
}
else{
printf("%d\n",ans[i]);
}
}
}

return 0;
}