BestCoder Round #62 (div.2) HDOJ5562 Clarke and food(脑洞)

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解决一个关于背包问题的变种，即如何最大化利用背包体积来装载不同大小的食物。通过排序和遍历策略找到能装下的最多食物数量。

Clarke and food

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 501 Accepted Submission(s): 292

Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke turned into a cook, was shopping for food.
Clarke has bought

n food. The volume of the

i th food is

vi . Now Clarke has a pack with volume

V . He wants to carry food as much as possible. Tell him the maxmium number he can brought with this pack.

Input

The first line contains an integer

T(1≤T≤10) , the number of the test cases.
For each test case:
The first line contains two integers

n,V(1≤n≤105,1≤V≤109) .
The second line contains

n integers, the

i th integer denotes

vi(1≤vi≤109) .

Output

For each test case, print a line with an integer which denotes the answer.

Sample Input

Sample Output

  
   2

Hint:  
We can carry 1 and 3, the total volume of them is 5.

题目链接:点击打开链接

脑洞题, 本以为是背包, 读入数据后排序, 从小到大一直拿, 拿到不能拿为止可以使装食物数量最大.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "utility"
#include "map"
#include "set"
#include "vector"
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 5;
int n, v, a[MAXN];
int main(int argc, char const *argv[])
{
    int t;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &v);
        for(int i = 0; i < n; ++i)
            scanf("%d", &a[i]);
        sort(a, a + n);
        int sum = 0, ans = 0;
        for(int i = 0; i < n; ++i) {
            sum += a[i];
            if(sum <= v) ans++;
            else break;
        }
        printf("%d\n", ans);
    }
    return 0;
}