C. Come to a spring outing

 

C. Come to a spring outing

Time Limit: 1000ms

Case Time Limit: 1000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

Submit Status PID: 29358

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    Spring is coming，WHU ACM team plans to organize a spring outing because of the good weather. They bought a lot of things to eat and use (collectively referred to the item), each item has a volume. In order to take them away, they also bought 3 backpacks, each backpack can be fitted to a certain volume of things, and the question is there: Can they take all the items away?

Input

    Input contains several test cases. The first line is an integer T indicates the number of test cases. For each test case, the first line is two integer n and m(1<=n<=30,1<=m<=400), indicate the number of items and the capacity of each backpack. The second line contains n integers, the i-th integer wi(1<=wi<=400) shows the volume of the i-th item.

Output

    For each set of data, output the case number first, and then output a string, “Yes” means they can take all things away, and “No” means they cannot take all things away by three backpack.

Sample Input

2
4 3
1 2 3 3
4 3
2 2 2 2

Sample Output

Case 1: Yes
Case 2: No

Submit Status PID: 29358

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int w[33];

int n,v;

bool cmp(int a,int b)

{

    return a>b;

}

bool OK=false;

void dfs(int i,int v1,int v2,int v3)

{

    if(i==n){ OK=true;  return; }

    if(OK==true) return;

    if(v1+w<=v)

        dfs(i+1,v1+w,v2,v3);

    if(v2+w<=v)

        dfs(i+1,v1,v2+w,v3);

    if(v3+w<=v)

        dfs(i+1,v1,v2,v3+w);

}

int main()

{

    int T;

    cin>>T;

    int cas=0;

while(T--)

{

    int sum=0;

    cin>>n>>v;

    for(int i=0;i<n;i++)

    {

        cin>>w;  sum+=w;

    }

    if(sum>3*v)

    {

        printf("Case %d: No\n",++cas);

        continue;

    }

    OK=false;

    sort(w,w+n,cmp);

    dfs(0,0,0,0);

    if(OK)

        printf("Case %d: Yes\n",++cas);

    else

        printf("Case %d: No\n",++cas);

}

    return 0;

}

转载于:https://www.cnblogs.com/CKboss/p/3350915.html