1408211934-hd-Crixalis's Equipment.cpp

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解决Crixalis（沙漠蝎子）搬家难题，通过合理的设备放置顺序确保所有装备能放入新家地洞。采用贪心算法策略，优先搬运空间效益比高的装备。

Crixalis's Equipment

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2817 Accepted Submission(s): 839

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

题目大意

沙漠里的皇帝--蝎子，要给自己换个洞，但是这个洞要足够大放得下自己的设备，现在给你t组数据，每组数据包含洞的体积v和蝎子的设备n，然后接下来的n行中一行包括两个数字v1(设备本身的体积),v2(移动设备所需要的体积)。如果洞的体积满足要求则输出Yes,否则输出No。

错误原因

一开始我想着根据v2做降序排列，然后依次递减作比较。但是WA了，事实证明我想的不够全面。

解题思路

我们反过来想我把 n 件物品全搬进洞，需要洞的最少体积是多少？

假设有两件物品 a（4，5）、b（6，8）先搬 a 再搬 b 则需的体积是 max(5,4+8)=12 ,反过来就是 max(8,6+5)=11

也就是说假设两件物品时a(x1,y1)、b(x2,y2)；答案就是 min (max(y1 , x1+y2) , max(y2 , x2+y1))

到这里就很自然的想比较 x1+y2 与 x2+y1 的大小假设 x1+y2 > x2+y1 则 y2-x2 > y1-x1 ，于是只要先搬 b 就得到 x2+y1，

同样如果 x1+y2 <= x2+y1 则 y2-x2 <= y1-x1 则需先搬 a；

于是我们的贪心策略是先搬差值大的，相等就随意！！

代码

#include<stdio.h>
#include<algorithm>
using namespace std;
struct hole
{
    int self,need,de;
}holes[1100];
/*
bool cmp(in a,in b)
{
    if(a.need!=b.need)
        return a.need>b.need;
    else
        return a.self>b.self;
}
   原来想着根据设备所要的体积大小排序。 */ 
bool cmp(hole a,hole b)
{
    return a.de>b.de;
}
/*假设有两件物品 a（4，5）、b（6，8）先搬 a 再搬 b 则需的体积是 max(5,4+8)=12 ,反过来就是 max(8,6+5)=11
也就是说假设两件物品时a(x1,y1)、b(x2,y2)；答案就是 min (max(y1 , x1+y2) , max(y2 , x2+y1))
 
到这里就很自然的想比较 x1+y2 与 x2+y1 的大小 假设 x1+y2 > x2+y1 则 y2-x2 > y1-x1 ，于是只要先搬 b 就得到 x2+y1，
同样如果 x1+y2 <= x2+y1 则 y2-x2 <= y1-x1 则需先搬 a；
 
于是我们的贪心策略是 先搬差值大的，相等就随意！！
 
int main()
{
    int t;
    int v,n;
    int i,j,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&v,&n);
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&holes[i].self,&holes[i].need);
            holes[i].de=holes[i].need-holes[i].self;
        }
        sort(holes,holes+n,cmp);
        j=0;
        for(i=0;i<n;i++)
        {
            if(v<holes[i].need)
                break;
            else
            {
                v-=holes[i].self;
                j++;
            }
        }
        if(j==n)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}