HDU Crixalis's Equipment

2023-11-03 06:10

文章标签 hdu crixalis equipment

本文主要是介绍HDU Crixalis's Equipment，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

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

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

Sample Output

  Yes
No

这也是一个贪心的问题，基本思路都一样。

题目大意：把n个物品放入洞穴中，输入放入后所占得体积Ai，和放入的过程中要用的体积Bi。问是否可以把物品全部放入洞穴中。

开始的时候就把放入时所需要的体积进行排序，再把放入后所占体积排序，那几个简单的测试数据过了，提交的时候是答案错误。

后面百度了一下，需要考虑的是俩个的体积差Bi-Ai,再对Ai排序。

AC代码：

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

struct node
{
int a,b,c;
}q[1008];

bool cmp(node x,node y)
{
if(x.c != y.c)
{
    return x.c>y.c;
}

    return x.a<y.a;
}
int main()
{
    int t,n,i,j,m;
    int ant;
    cin>>t;
    while(t--)
    {
    ant=0;
        cin>>n>>m;
        for(i=0;i<m;i++) {
    cin>>q[i].a>>q[i].b;
    q[i].c=q[i].b-q[i].a;
        }

        sort(q,q+m,cmp);
        for(i=0;i<m;i++)
        {
        if(q[i].b<= n)
        {
          n=n-q[i].a;
          ant++;
        }
        else
           break;
        }
        if(i < m)
          cout<<"No"<<endl;
        else
          cout<<"Yes"<<endl;
    }
    return 0;
}