zoukankan      html  css  js  c++  java
  • Crixalis's Equipment

    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
    AC代码:

    #include <iostream>
    #include
    <stdlib.h>
    using namespace std;
    int flag;
    struct equip
    {
    int a;
    int b;
    int c;
    } equips[
    1002];
    void moveEquip(int i,int v,int n)
    {
    if (v<0)return;

    if(equips[i].b>v)
    {
    flag
    =0;
    }
    else
    {
    if(n==1)return ;
    v
    -=equips[i].a;
    moveEquip(
    ++i,v,--n);
    }
    }
    int comp(const void *a,const void *b)
    {
    struct equip *Ea=(struct equip *)a;
    struct equip *Eb=(struct equip *)b;
    if(Ea->c!=Eb->c)
    {
    return Eb->c-Ea->c;
    }
    else
    {
    return Ea->a-Eb->a;
    }
    }

    int main(void)
    {
    int t;
    cin
    >>t;
    while (t>0)
    {
    t
    --;
    int i,v,n;
    cin
    >>v>>n;

    for(i=0; i<n; i++)
    {
    cin
    >>equips[i].a>>equips[i].b;
    equips[i].c
    =equips[i].b-equips[i].a;
    }
    flag
    =1;
    qsort(equips,n,
    sizeof(equips[0]),comp);
    moveEquip(
    0,v,n);
    if(flag==1)
    {
    cout
    <<"Yes"<<endl;
    }
    else
    {
    cout
    <<"No"<<endl;
    }
    }
    return 0;
    }
  • 相关阅读:
    Linux下hook指定库
    一行一行往上爬
    高可用数据同步方案-SqlServer迁移Mysql实战
    Hystrix核心基础
    Fastjson解析多级泛型的几种方式—使用class文件来解析多级泛型
    面试大全之JVM篇
    云原生下的CICD
    学习Raft算法的笔记
    Go语言下的线程模型
    分布式事务解决方案以及 .Net Core 下的实现(上)
  • 原文地址:https://www.cnblogs.com/aboutblank/p/2120750.html
Copyright © 2011-2022 走看看