Crixalis‘s Equipment |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 3097 Accepted Submission(s): 922 |
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Problem Description
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 |
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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 0<T<= 10, 0<V<10000, 0<N<1000, 0 <Ai< V, Ai <= Bi < 1000. |
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Output For each case output "Yes" if Crixalis can move all his equipment into the new hole or else output "No". |
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Sample Input 2 20 3 10 20 3 10 1 7 10 2 1 10 2 11 |
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Sample Output Yes No |
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Source HDU 2009-10 Programming Contest |
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Recommend lcy |
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分析参考：http://www.cnblogs.com/Action-/archive/2012/07/03/2575069.html
#include<stdio.h>
#include<stdlib.h>
struct Equipment
{
int ai;
int bi;
}ep[1000];
int comp(const void *a,const void *b)
{
struct Equipment *c=(struct Equipment *)a;//类型转换
struct Equipment *d=(struct Equipment *)b;
return (c->ai+d->bi) - (c->bi+d->ai);//不能用大于号代替减号，否则不能AC
}
int main()
{
int num=0;
int V=0,N=0;
int i=0;
int flag=0;//能否成功的标志。1为Yes
scanf("%d",&num);
while(num--)
{
flag=1;
scanf("%d%d",&V,&N);
for(i=0;i<N;i++)
{
scanf("%d%d",&ep[i].ai,&ep[i].bi);
}
qsort(ep,N,sizeof(struct Equipment),comp);
for(i=0;i<N;i++)
{
if(ep[i].bi>V)
{
flag=0;
break;
}
V-=ep[i].ai;
}
if(flag==0)
printf("No\n");
else
printf("Yes\n");
}
return 0;
}
Crixalis's Equipment