Crixalis‘s Equipment
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2562 Accepted Submission(s): 1056
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
分析：
停放体积 移动体积
第一件物品 a1 b1
第二件物品 a2 b2
如果这两件物品的移动体积都不大于洞的体积V
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中比較小的先放.
如果n件物品的移动体积都不大于洞的体积V(如果有大于的 那么结果必定是NO)
将N件物品依照a1+b2<a2+b1进行排序,然后依次放入洞中。即依照差值从大到小放入洞中.
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
int x,y;
}p[10005];
bool cmp(node a,node b) // 差值排序
{
return (a.y-a.x)>(b.y-b.x);
}
int main ()
{
int t,n,m,i,j;
int flag;
scanf("%d",&t);
for(i=0;i<t;i++)
{
scanf("%d%d",&n,&m);
for(j=0;j<m;j++)
scanf("%d%d",&p[j].x,&p[j].y);
sort(p,p+m,cmp);
flag=1;
for(j=0;j<m;j++)
{
if(p[j].y>n)
{
flag=0;
break;
}
n-=p[j].x;
}
if(flag) printf("Yes\n");
else printf("No\n");
}
return 0;
}
