HDU 3177 Crixalis's Equipment(差排)
Crixalis's Equipment
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1429 Accepted Submission(s): 583
Problem Description
![](http://acm.hdu.edu.cn/data/images/C241-1003-1.jpg)
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.
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
Recommend
lcy
分析:(分析转:http://www.cnblogs.com/Action-/archive/2012/07/03/2575069.html)
该题讲的是搬家问题,搬运东西需要占多大空间还有搬运时要求的空间。该题的贪心原则则是尽可能的装入东西,即
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,即按照差值从大到小放入洞中
#include<iostream> #include<algorithm> using namespace std; struct Edge { int a,b; }edge[2000]; int cmp(Edge A,Edge B)//差排!!! { return A.a+B.b<B.a+A.b; /*将n件物品按照a1+b2<a2+b1进行排序,然后依次放入洞中,即按照差值从大到小放入洞中, 选择a1+b2和a2+b1中比较小的先放入*/ } int main() { int t,v,n,mark,sum,i; scanf("%d",&t); while(t--) { scanf("%d %d",&v,&n); sum=v; mark=1; for(i=0;i<n;i++) scanf("%d %d",&edge[i].a,&edge[i].b); sort(edge,edge+n,cmp); for(i=0;i<n;i++) { if(edge[i].b>sum) { mark=0; break; } else { sum-=edge[i].a; mark=1; } } if(mark) printf("Yes\n"); else printf("No\n"); } return 0; }