HDU 3177 Crixalis's Equipment

http://acm.hdu.edu.cn/showproblem.php?pid=3177

Crixalis's Equipment

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 594    Accepted Submission(s): 229

Problem Description

Crixalis - Sand King used to be a giant scorpion(蝎子) in the deserts of Kalimdor. Though he's a guardian of Lich King now, he keeps the living habit of a scorpion like living underground and digging holes.

Someday Crixalis decides to move to another nice place and build a new house for himself (Actually it's just a new hole). As he collected a lot of equipment, he needs to dig a hole beside his new house to store them. This hole has a volume of V units, and Crixalis has N equipment, each of them needs Ai units of space. When dragging his equipment into the hole, Crixalis finds that he needs more space to ensure everything is placed well. Actually, the ith equipment needs Bi units of space during the moving. More precisely Crixalis can not move equipment into the hole unless there are Bi units of space left. After it moved in, the volume of the hole will decrease by Ai. Crixalis wonders if he can move all his equipment into the new hole and he turns to you for help.

 

Input

The first line contains an integer T, indicating the number of test cases. Then follows T cases, each one contains N + 1 lines. The first line contains 2 integers: V, volume of a hole and N, number of equipment respectively. The next N lines contain N pairs of integers: Ai and Bi.
0<T<= 10, 0<V<10000, 0<N<1000, 0 <Ai< V, Ai <= Bi < 1000.

 

Output

For each case output "Yes" if Crixalis can move all his equipment into the new hole or else output "No".

 

Sample Input

2 20 3 10 20 3 10 1 7 10 2 1 10 2 11

 

Sample Output

Yes No

 

Source

HDU 2009-10 Programming Contest

 

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题目大意：向一个容量为V的洞中搬物品 每件物品有一个停放体积 可一个移动体积 问能否放下这些物品

解题思路：对这些物品进行排序 按照顺序依次进入洞中 排序要尽可能使得所有的东西都能进入洞中

这是一个贪心的问题  

                                   停放体积         移动体积

                  第一件物品          a1             b1

 

                  第二件物品          a2             b2

假设这两件物品的移动体积都不大于洞的体积V

那么将单独比较两个物品的时候会发现  a1+b2为先放第一件物品 后放第二件物品的最大瞬时体积

                                                  a2+b1为先放第二件物品 后放第一件物品的最大瞬时体积

我们应该选择a1+b2和a2+b1中比较小的先放

那么从2件物品 扩展到N件物品 假设n件物品的移动体积都不大于洞的体积V(如果有大于的 那么结果必然是NO)

将N件物品按照a1+b2<a2+b1进行排序 然后依次放入洞中

#include<iostream>
#include<algorithm>
using namespace std;

struct Node
{
    int f,l;
}a[1001];

bool cmp(Node x,Node y)
{
    return x.f+y.l<y.f+x.l;//排序方式是重点
}

int main()
{
    int t;
    cin>>t;
    for(int i=1;i<=t;i++)
    {
        int v,n;
        cin>>v>>n;
        for(int j=1;j<=n;j++)
        {
            cin>>a[j].f>>a[j].l;
        }
        sort(a+1,a+n+1,cmp);
        int k=0;
        for(int j=1;j<=n;j++)
        {
            if(a[j].l>v)
            {
                cout<<"No"<<endl;
                k=1;
                break;
            }
            else
            v-=a[j].f;
        }
        if(!k)
        cout<<"Yes"<<endl;

    }
    return 0;
}