Crixalis's Equipment
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3092 Accepted Submission(s): 1255
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 scorpionlike 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.
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.
2 20 3 10 20 3 10 1 7 10 2 1 10 2 11Sample Output
/*
Crixalis's Equipment 给定一个最大空间和N个数量体积
要求放入N个 每个都有实际占有的空间 以及放入最少需要占有的时间
*/
#include<iostream>
#include<algorithm>
using namespace std;
struct volume{
int vol;
int leastvol;
};
bool cmp(volume a,volume b)
{
return a.leastvol-a.vol>b.leastvol-b.vol;
}
int main(){
volume a[1001];
int m,n,i,v;
scanf("%d",&m);
while(m--)
{
scanf("%d%d",&v,&n);
for(i=0;i<n;i++)
scanf("%d%d",&a[i].vol,&a[i].leastvol);
sort(a,a+n,cmp);
for(i=0;i<n;i++)
{
if(v>=a[i].leastvol) {
v-=a[i].vol;
}
else {
v=-1;
break;
}
}
if(v>=0) printf("Yes\n");
else printf("No\n");
}
return 0;
}